www.vnmath.com 1 BI DNG HC SINH GII TON 9 www.vnmath.comwww.VNMATH.comwww.vnmath.com 2 CHUYN 1 : A THC B. CC PHNG PHP V BI TP: I. TCH MT HNG T THNH NHIU HNG T: * nh l b sung: + a thc f(x) c nghim hu t th c dng p/q trong p l c ca h s t do, q l c dng ca h s cao nht+ Nu f(x) c tng cc h s bng 0 th f(x) c mt nhn t l x 1 + Nu f(x) c tng cc h s ca cc hng t bc chn bng tng cc h s ca cc hng t bc l th f(x) c mt nhn t l x + 1 + Nu a l nghim nguyn ca f(x) v f(1); f(- 1) khc 0 th f(1)a - 1 v f(-1)a + 1 u l s nguyn. nhanh chng loi tr nghim l c ca h s t do 1. V d 1: 3x2 8x + 4 Cch 1: Tch hng t th 2 3x2 8x + 4 =3x2 6x 2x+ 4 = 3x(x 2) 2(x 2) = (x 2)(3x 2) Cch 2: Tch hng t th nht: 3x2 8x + 4 =(4x2 8x+ 4)- x2 = (2x 2)2 x2 = (2x 2 + x)(2x 2 x)= (x 2)(3x 2) 2. V d 2: x3 x2 - 4 Ta nhn thy nghim ca f(x) nu c th x =1; 2; 4 , ch c f(2) = 0 nn x = 2 l nghimca f(x) nn f(x) c mt nhn t l x 2. Do tatch f(x) thnh cc nhm c xut hin mt nhn t l x 2 Cch 1: x3 - x2 4 = ( ) ( ) ( ) ( )3 2 2 2x 2x x 2x 2x 4 x x 2 x(x 2) 2(x 2) + + = + + = ( )( )2x 2 x x 2 + + Cch 2: ( ) ( )3 2 3 2 3 2x x 4 x 8 x 4 x 8 x 4 = + = 2(x 2)(x 2x 4) (x 2)(x 2) = + + + = ( ) ( )2 2x 2 x 2x 4 (x 2) (x 2)(x x 2) ( + + + = + + 3. V d 3:f(x) =3x3 7x2 + 17x 5 Nhn xt:1, 5 khng l nghim ca f(x), nh vy f(x) khngc nghim nguyn. Nn f(x) nu c nghim th l nghim hu t Ta nhn thy x = 13 l nghim ca f(x) do f(x) c mt nhn t l3x 1. Nn f(x) =3x3 7x2 + 17x 5 = ( ) ( ) ( )3 2 2 3 2 23x x 6x 2x 15x 5 3x x 6x 2x 15x 5 + + = + = 2 2x (3x 1) 2x(3x 1) 5(3x 1) (3x 1)(x 2x 5) + = +V 2 2 2x 2x 5 (x 2x 1) 4 (x 1) 4 0 + = + + = + >vi mi x nn khng phn tch c thnh nhn t na 4. V d 4: x3 + 5x2 + 8x+ 4Nhn xt: Tng cc h s ca cc hng t bc chn bng tng cc h s ca cc hng t bc l nn a thc c mt nhn t l x + 1 x3 + 5x2 + 8x+ 4 = (x3 + x2 ) + (4x2 + 4x) + (4x + 4) = x2(x + 1) + 4x(x + 1) + 4(x + 1) www.VNMATH.comwww.vnmath.com 3= (x + 1)(x2 + 4x + 4) = (x + 1)(x + 2)2 5. V d 5:f(x) = x5 2x4 + 3x3 4x2 + 2 Tng cc h s bng 0 th nn a thc c mt nhn t l x 1, chia f(x) cho (x 1) ta c: x5 2x4 + 3x3 4x2 + 2 = (x 1)(x4- x3+ 2 x2 - 2 x - 2) V x4- x3+ 2 x2 - 2 x - 2khng c nghim nguyn cng khng c nghim hu t nn khng phn tch c na 6.V d 6: x4 + 1997x2 + 1996x + 1997 = (x4 + x2 + 1) + (1996x2 + 1996x + 1996) =(x2 + x+ 1)(x2 - x+ 1) + 1996(x2 + x+ 1)=(x2 + x+ 1)(x2 - x+ 1 + 1996)=(x2 + x+ 1)(x2 - x+ 1997) 7. V d 7:x2 -x - 2001.2002 = x2 -x - 2001.(2001 + 1) = x2 -x 20012 - 2001 = (x2 20012) (x + 2001) = (x + 2001)(x 2002) II. THM , BT CNG MT HNG T: 1. Thm, bt cng mt s hng t xut hin hiu hai bnh phng: a) V d 1: 4x4 + 81 = 4x4+ 36x2 + 81 - 36x2 = (2x2 + 9)2 36x2

= (2x2 + 9)2 (6x)2 = (2x2 + 9 + 6x)(2x2 + 9 6x)= (2x2 + 6x + 9 )(2x2 6x + 9)b) V d 2: x8 + 98x4 + 1 = (x8 + 2x4 + 1 ) + 96x4

= (x4 + 1)2 + 16x2(x4 + 1) + 64x4 - 16x2(x4 + 1) + 32x4 = (x4 + 1 + 8x2)2 16x2(x4 + 1 2x2) = (x4 + 8x2 + 1)2- 16x2(x2 1)2 = (x4 + 8x2 + 1)2- (4x3 4x )2

= (x4 + 4x3 + 8x2 4x + 1)(x4 - 4x3 + 8x2 + 4x + 1) 2. Thm, bt cng mt s hng t xut hin nhn t chung a) V d 1: x7 + x2 + 1 = (x7 x)+ (x2 + x + 1 ) =x(x6 1) + (x2 + x + 1 )=x(x3 - 1)(x3 + 1) + (x2 + x + 1 ) = x(x 1)(x2 + x + 1 ) (x3 + 1) + (x2 + x + 1) =(x2 + x + 1)[x(x 1)(x3 + 1) + 1] = (x2 + x + 1)(x5 x4+ x2- x + 1) b) V d 2: x7 + x5 + 1 = (x7 x ) + (x5 x2 ) + (x2 + x + 1)= x(x3 1)(x3 + 1) + x2(x3 1) + (x2 + x + 1)= (x2 + x + 1)(x 1)(x4 + x) + x2 (x 1)(x2 + x + 1) + (x2 + x + 1)= (x2 + x + 1)[(x5 x4 + x2 x) + (x3 x2 ) + 1] = (x2 + x + 1)(x5 x4 + x3 x + 1)* Ghi nh:Cc a thc c dng x3m + 1 + x3n + 2 + 1 nh: x7 + x2 + 1 ; x7 + x5 + 1 ; x8 + x4 + 1 ; x5 + x + 1 ; x8 + x + 1 ; u c nhn t chung lx2 + x + 1 III. T BIN PH: 1. V d 1:x(x + 4)(x + 6)(x + 10) + 128 = [x(x + 10)][(x + 4)(x + 6)] + 128 =(x2 + 10x) + (x2 + 10x+ 24) + 128 tx2 + 10x + 12 =y, a thc c dng (y 12)(y + 12) + 128 = y2 144 + 128 = y2 16 = (y + 4)(y 4) =( x2 + 10x + 8 )(x2+ 10x+ 16 ) =(x + 2)(x + 8)( x2 + 10x + 8 ) 2. V d 2:A = x4 + 6x3 + 7x2 6x + 1 Gi s x=0 ta vitx4 + 6x3 + 7x2 6x + 1 =x2 ( x2 + 6x + 7 26 1 + x x) = x2 [(x2 + 21 x) + 6(x - 1 x) + 7 ] tx - 1 x = ythx2 + 21 x = y2 + 2, do www.VNMATH.comwww.vnmath.com 4A = x2(y2 + 2 + 6y + 7) = x2(y + 3)2=(xy + 3x)2 = [x(x - 1 x)2 + 3x]2 = (x2 + 3x 1)2

* Ch : V d trn c th gii bng cch p dng hng ng thc nh sau: A = x4 + 6x3 + 7x2 6x + 1 = x4 + (6x3 2x2 ) + (9x2 6x + 1 ) =x4 + 2x2(3x 1) + (3x 1)2 = (x2 + 3x 1)2

3. V d 3:A = 2 2 2 2 2(x y z )(x y z) (xy yz+zx) + + + + + += 2 2 2 2 2 2 2(x y z ) 2(xy yz+zx) (x y z ) (xy yz+zx)( + + + + + + + + t2 2 2x y z + += a, xy + yz + zx = b ta cA =a(a + 2b) + b2 = a2 + 2ab + b2 = (a + b)2 =( 2 2 2x y z + ++ xy + yz + zx)2 4. V d 4: B = 4 4 4 2 2 2 2 2 2 2 2 42( ) ( ) 2( )( ) ( ) x y z x y z x y z x y z x y z + + + + + + + + + + +tx4 + y4 + z4 = a,x2 + y2 + z2 = b, x + y + z = c ta c: B = 2a b2 2bc2 + c4 = 2a 2b2+ b2 - 2bc2 + c4 = 2(a b2) + (b c2)2 Ta li c: a b2 =- 2(2 2 2 2 2 2xy yz zx + + ) v b c2 = - 2(xy + yz + zx) Do : B = - 4(2 2 2 2 2 2x y y z z x + + ) + 4 (xy + yz + zx)2

2 2 2 2 2 2 2 2 2 2 2 2 2 2 24x y 4y z 4z x 4x y 4y z 4z x 8x yz 8xy z 8xyz8xyz(x y z)= + + + + + += + + 5. V d 5:3 3 3 3(a b c) 4(a b c ) 12abc + + + + t a + b = m, a b = nth 4ab = m2 n2 a3 + b3 = (a + b)[(a b)2 + ab] = m(n2 + 2 2m- n4). Ta c: C = (m + c)3 4. 3 23 2 2m+ 3mn4c 3c(m-n )4 = 3( - c3 +mc2 mn2 + cn2) = 3[c2(m - c) - n2(m - c)] = 3(m - c)(c - n)(c + n) = 3(a + b - c)(c + a - b)(c - a + b) IV. PHNG PHP H S BT NH: 1. V d 1: x4 - 6x3 + 12x2 - 14x + 3 Nhn xt: cc s 1, 3 khng l nghim ca a thc, a thc khng c nghim nguyn cng khng c nghim hu t Nh vy nu a thc phn tch c thnh nhn t th phi c dng(x2 + ax + b)(x2 + cx + d) = x4 + (a + c)x3 + (ac + b + d)x2 + (ad + bc)x + bd ng nht a thc ny vi a thc cho ta c: a c 6ac b d 12ad bc 14bd 3+ = + + =+ = = Xt bd = 3 vib, d e Z, b e { } 1, 3 vi b = 3 th d = 1 h iu kin trn tr thnh a c 6ac 8 2c 8 c 4a 3c 14 ac 8 a 2bd 3+ = = = = + = = = = Vy: x4 - 6x3 + 12x2 - 14x + 3 =(x2 - 2x + 3)(x2 - 4x+ 1)www.VNMATH.comwww.vnmath.com 52. V d 2: 2x4 - 3x3 - 7x2 + 6x + 8 Nhn xt: a thc c 1 nghim l x = 2 nn c tha s lx - 2 do ta c: 2x4 - 3x3 - 7x2 + 6x + 8 = (x - 2)(2x3 + ax2 + bx + c)=2x4 + (a - 4)x3 + (b - 2a)x2 + (c - 2b)x - 2c a 4 3a 1b 2a 7b 5c 2b 6c 42c 8 = = = = = = =

Suy ra:2x4 - 3x3 - 7x2 + 6x + 8 = (x - 2)(2x3 + x2 - 5x- 4)Ta li c 2x3 + x2 - 5x- 4 l a thc c tng h s ca cc hng t bc l v bc chn bng nhau nn c 1 nhn t lx + 1 nn2x3 + x2 - 5x- 4 = (x + 1)(2x2 - x - 4) Vy: 2x4 - 3x3 - 7x2 + 6x + 8 = (x - 2)(x + 1)(2x2 - x - 4) 3. V d 3:12x2 + 5x - 12y2 + 12y - 10xy - 3 = (a x + by + 3)(cx + dy- 1) =acx2 + (3c - a)x+ bdy2 + (3d - b)y + (bc + ad)xy 3ac 12a 4bc ad 10c 33c a 5b 6bd 12d 23d b 12= =+ = = = = = = = 12x2 + 5x - 12y2 + 12y - 10xy - 3 = (4 x - 6y + 3)(3x + 2y- 1) BI TP:Phn tch cc a thc sau thnh nhn t: www.vnmath.com 1) x3 - 7x + 6 2) x3 - 9x2 + 6x + 16 3) x3 - 6x2 - x + 30 4) 2x3 - x2 + 5x + 3 5) 27x3 - 27x2 + 18x - 4 6) x2 + 2xy + y2- x - y - 12 7) (x + 2)(x +3)(x + 4)(x + 5) - 24 8) 4x4 - 32x2 + 1 9) 3(x4 + x2 + 1) - (x2 + x + 1)2 10)64x4 + y4 11) a6 + a4 + a2b2 + b4 - b6 12) x3 + 3xy + y3 - 1 13) 4x4 + 4x3 + 5x2 + 2x + 1 14)x8 + x + 1 15) x8 + 3x4 + 416) 3x2 + 22xy + 11x + 37y + 7y2 +10 17) x4 - 8x + 63 www.VNMATH.comwww.vnmath.com 6 CHUYN 2- LU THA BC N CA MT NH THC B. KIN THC V BI TP VN DNG: I. Mt s hng ng thc tng qut: 1.an - bn= (a - b)(an - 1 + an - 2 b + an - 3 b2 + + abn - 2 + bn - 1 ) 2.an + bn= (a + b) ( an - 1 - an - 2b+ an - 3b2 - - abn - 2 + bn - 1 ) 3. Nh thc Niutn: (a + b)n= an + 1nC an - 1 b + 2nC an - 2 b2 + +n1nCab n - 1 + bn Trong : k nn(n - 1)(n - 2)...[n - (k - 1)]C1.2.3...k=: T hp chp k ca n phn t II. Cch xc nh h s ca khai trin Niutn: 1. Cch 1: Dng cng thc k nn(n - 1)(n - 2)...[n - (k - 1)]Ck !=Chng hn h s ca hng ta4b3 trong khai trin ca (a + b)7 l 477.6.5.4 7.6.5.4C354! 4.3.2.1= = =Ch :a) k nn !C n!(n - k) !=vi quy c0! = 1 477! 7.6.5.4.3.2.1C354!.3! 4.3.2.1.3.2.1= = =b) Ta c: k nC = k - 1 nC nn 4 37 77.6.5.CC353!= = = 2. Cch 2: Dng tam gic Patxcan nh1 Dng 1(n = 1)11 Dng 2(n = 1)121 Dng 3(n = 3)1331 Dng 4(n = 4)14641 Dng 5(n = 5)15101051 Dng 6(n = 6)161520 15 61 Trong tam gic ny, hai cnh bn gm cc s 1; dng k + 1 c thnh lp t dng k (k>1), chng hn dng 2 (n = 2) ta c 2 = 1 + 1, dng 3 (n = 3): 3 = 2 + 1, 3 = 1 + 2dng 4 (n = 4): 4 = 1 + 3, 6 = 3 + 3, 4 = 3 + 1, Vi n = 4 th: (a + b)4 = a4 + 4a3b + 6a2b2 + 4ab3 + b4 Vi n = 5 th: (a + b)5 = a5 + 5a4b + 10a3b2 + 10a2b3 + 5ab4 + b5 Vi n = 6 th:(a + b)6 = a6 + 6a5b + 15a4b2 + 20a3b3 + 15a2 b4 + 6ab5 + b6 3. Cch 3: Tm h s ca hng t ng sau theo cc h s ca hng t ng trc: a) H s ca hng t th nht bng 1 b) Mun c h s ca ca hng tth k + 1, ta ly h s ca hng t th k nhn vi s m ca bin trong hng t th k ri chia cho kChng hn: (a + b)4

= a4 + 1.41a3b + 4.32a2b2 + 4.3.22.3 ab3 + 4.3.2.2.3.4 b5 Ch rng: cc h s ca khai trin Niutn c tnh i xng qua hng t ng gia, nghal cc hng t cch u hai hng t u v cui c h s bng nhau (a + b)n = an + nan -1b + n(n - 1)1.2an - 2b2 + + n(n - 1)1.2a2bn- 2 + nan - 1bn - 1 + bn www.VNMATH.comwww.vnmath.com 7III. V d: 1. V d 1: phn tch a thc sau thnh nhn ta) A =(x + y)5 - x5 - y5 Cch 1: khai trin (x + y)5 ri rt gn A A=(x + y)5 - x5 - y5 = ( x5 + 5x4y + 10x3y2+ 10x2y3 + 5xy4 + y5) - x5 - y5 =5x4y + 10x3y2+ 10x2y3 + 5xy4 = 5xy(x3 + 2x2y + 2xy2 + y3) =5xy [(x + y)(x2 - xy + y2) + 2xy(x + y)] = 5xy(x + y)(x2 + xy + y2) Cch 2: A = (x + y)5 - (x5 + y5) x5 + y5 chia ht cho x+ ynn chia x5 + y5 cho x + y ta c:x5 + y5 = (x + y)(x4 - x3y + x2y2 - xy3 + y4)nn A c nhn t chung l (x + y), t (x + y) lm nhn t chung, ta tm c nhn t cn li b) B = (x + y)7 - x7 - y7 = (x7+7x6y +21x5y2 + 35x4y3 +35x3y4 +21x2y57xy6 + y7) - x7 - y7

= 7x6y + 21x5y2 + 35x4y3 + 35x3y4 + 21x2y5 + 7xy6

=7xy[(x5 + y5 ) + 3(x4y

+ xy4) + 5(x3y2 + x2y3 )] = 7xy {[(x + y)(x4 - x3y + x2y2 - xy3 + y4) ] + 3xy(x + y)(x2 - xy + y2) + 5x2y2(x + y)} = 7xy(x + y)[x4 - x3y + x2y2 - xy3 + y4 + 3xy(x2 + xy + y2) + 5x2y2 ] = 7xy(x + y)[x4 - x3y + x2y2 - xy3 + y4 + 3x3y - 3x2y2 + 3xy3 + 5x2y2 ] = 7xy(x + y)[(x4 + 2x2y2 + y4) + 2xy (x2 + y2) + x2y2 ] =7xy(x + y)(x2 + xy + y2 )2 V d 2:Tm tng h s cc a thc c c sau khi khai trin a) (4x - 3)4 Cch 1: Theo cnh thc Niu tn ta c: (4x - 3)4 = 4.(4x)3.3 + 6.(4x)2.32 - 4. 4x. 33 + 34 = 256x4 - 768x3 + 864x2 - 432x + 81 Tng cc h s: 256 - 768+ 864 - 432 + 81 = 1 b) Cch 2:Xt ng thc (4x - 3)4 = c0x4 + c1x3 + c2x2 + c3x + c4 Tng cc h s:c0 + c1 + c2 + c3 + c4 Thay x = 1 vo ng thc trn ta c: (4.1 - 3)4 =c0 + c1 + c2 + c3 + c4 Vy: c0 + c1 + c2 + c3 + c4 = 1 * Ghi ch: Tng cc h s khai trin ca mt nh thc, mt a thc bng gi tr ca athc ti x = 1 C. BI TP: Bi 1: Phn tch thnh nhn t a) (a + b)3 - a3 - b3 b) (x + y)4 + x4 + y4 Bi 2: Tm tng cc h s c c sau khi khai trin a thc a) (5x - 2)5b) (x2+ x - 2)2010 + (x2 - x + 1)2011 www.vnmath.com www.VNMATH.comwww.vnmath.com 8 CHUYN 3 - CC BI TON V S CHIA HT CA S NGUYN B.KIN THC V CC BI TON: I. Dng 1: Chng minh quan h chia ht 1. Kin thc: * chng minh A(n) chia ht cho mt s m ta phn tch A(n) thnh nhn t c mt nhn t lm hoc bi ca m, nu m l hp s th ta li phn tch n thnh nhn t c cc oi mt nguyn t cng nhau, ri chng minh A(n) chia ht cho cc s * Ch : + Vi k s nguyn lin tip bao gi cng tn ti mt bi ca k + Khi chng minh A(n) chia ht cho m ta xt mi trng hp v s d khi chia A(n) cho m + Vi mi s nguyn a, b v s t nhin n th: 2. Bi tp: 2. Cc bi ton Bi 1: chng minh rng a) 251 - 1 chia ht cho 7 b) 270 + 370 chia ht cho 13 c) 1719 + 1917 chi ht cho 18d) 3663 - 1 chia ht cho 7 nhng khng chia ht cho 37 e) 24n-1 chia ht cho 15 vi ne NGii a) 251 - 1 = (23)17 - 1 23 - 1 = 7 b) 270 + 370 (22)35 + (32)35 = 435 + 935 4 + 9 = 13 c) 1719 + 1917 =(1719 + 1) + (1917 - 1)1719 + 1 17 + 1 = 18 v 1917 - 1 19 - 1 = 18 nn(1719 + 1) + (1917 - 1)hay 1719 + 1917 18 d) 3663 - 1 36 - 1 = 35 7 3663 - 1 = (3663 + 1) - 2chi cho 37 d - 2 e) 2 4n - 1 = (24) n - 1 24 - 1 = 15 Bi 2: chng minh rng a)n5 - n chia ht cho 30 vi n e N; b) n4 -10n2+ 9 chia ht cho 384 vi mi n l neZ c) 10n

+18n -28 chia ht cho 27 vi ne N; Gii: a) n5 - n = n(n4 - 1) = n(n - 1)(n + 1)(n2 + 1) = (n - 1).n.(n + 1)(n2 + 1) chia ht cho 6 v (n - 1).n.(n+1) l tch ca ba s t nhin lin tip nn chia ht cho 2 v 3 (*) Mt khc n5 - n = n(n2 - 1)(n2 + 1) = n(n2 - 1).(n2 - 4 + 5) = n(n2 - 1).(n2 - 4 ) + 5n(n2 - 1) = (n - 2)(n - 1)n(n + 1)(n+ 2) + 5n(n2 - 1) V (n - 2)(n - 1)n(n + 1)(n+ 2) l tch ca 5 s t nhin lin tip nn chia ht cho 5 +) an - bnchia ht cho a - b (a = - b) +) a2n + 1 + b2n + 1 chia ht cho a + b + (a + b)n = B(a) + bn +) (a+ 1)n l BS(a )+ 1 +)(a - 1)2n l B(a) + 1 +) (a - 1)2n + 1 l B(a) - 1 www.VNMATH.comwww.vnmath.com 9 5n(n2 - 1) chia ht cho 5 Suy ra (n - 2)(n - 1)n(n + 1)(n+ 2) + 5n(n2 - 1) chia ht cho 5 (**) T (*) v (**) suy ra pcm b) t A = n4 -10n2+ 9 = (n4 -n2 ) - (9n2 - 9) =(n2 - 1)(n2 - 9) = (n - 3)(n - 1)(n + 1)(n + 3) V n l nn t n = 2k + 1 (k e Z) thA = (2k - 2).2k.(2k + 2)(2k + 4) = 16(k - 1).k.(k + 1).(k + 2) A chia ht cho 16 (1) V(k - 1).k.(k + 1).(k + 2) l tch ca 4 s nguyn lin tip nn A c cha bi ca 2, 3, 4 nn A l bi ca 24 hay A chia ht cho 24 (2) T (1) v (2) suy ra A chia ht cho 16. 24 = 384 c) 10 n

+18n -28 =( 10 n - 9n - 1) + (27n - 27) + Ta c: 27n - 27 27 (1) + 10 n - 9n - 1 = [(n9...9 + 1) - 9n - 1] =n9...9 - 9n= 9( n1...1 - n) 27 (2) v 9 9 v n1...1 - n 3 do n1...1 - nl mt s c tng cc ch s chia ht cho 3T (1) v (2) suy ra pcm3. Bi 3: Chng minh rng vi mi s nguyn a th a) a3 - achia ht cho 3 b) a7 - achia ht cho 7 Gii a) a3 - a= a(a2 - 1) =(a - 1) a (a + 1)l tch ca ba s nguyn lin tip nn tn ti mt s l bi ca 3 nn(a - 1) a (a + 1) chia ht cho 3 b) ) a7 - a= a(a6 - 1) = a(a2 - 1)(a2 + a + 1)(a2 -a + 1) Nu a = 7k (k e Z) th a chia ht cho 7 Nu a = 7k + 1 (k eZ)th a2 - 1 = 49k2 + 14kchia ht cho 7 Nu a = 7k + 2 (k eZ)th a2 + a + 1 = 49k2 + 35k+ 7 chia ht cho 7 Nu a = 7k + 3 (k eZ)th a2 - a + 1 = 49k2 + 35k+ 7 chia ht cho 7 Trong trng hp no cng c mt tha s chia ht cho 7 Vy: a7 - achia ht cho 7 Bi 4: Chng minh rngA = 13 + 23 + 33 + ...+ 1003 chia ht cho B = 1 + 2 + 3 + ... + 100 Gii Ta c: B = (1 + 100) + (2 + 99) + ...+ (50 + 51) = 101. 50 chng minh A chia ht cho B ta chng minh A chia ht cho 50 v 101 Ta c: A = (13 + 1003) + (23 + 993) + ... +(503 + 513)= (1 + 100)(12 + 100 + 1002) + (2 + 99)(22 + 2. 99 + 992) + ... + (50 + 51)(502 + 50. 51 + 512) = 101(12 + 100 + 1002 + 22 + 2. 99 + 992 + ... + 502 + 50. 51 + 512) chia ht cho 101 (1) Li c:A = (13 + 993) + (23 + 983) + ... + (503 + 1003) Mi s hng trong ngoc u chia ht cho 50 nn A chia ht cho 50 (2)T (1) v (2) suy ra A chia ht cho 101 v 50 nn A chi ht cho B Bi tp v nh Chng minh rng: a) a5 a chia ht cho 5 b) n3 + 6n2 + 8n chia ht cho 48 vi mi n chn c) Cho a l s nguyn t ln hn 3. Cmra2 1 chia ht cho 24 www.VNMATH.comwww.vnmath.com 10d) Nu a + b + c chia ht cho 6 th a3 + b3 + c3 chia ht cho 6 e) 20092010khng chia ht cho 2010 f) n2 + 7n + 22khng chia ht cho 9 Dng 2: Tm s d ca mt php chia Bi 1:Tm s d khi chia 2100

a)cho 9, b) cho 25, c) cho 125 Gii a) Lu tha ca 2 st vi bi ca 9 l 23 = 8 = 9 - 1 Ta c : 2100 = 2. (23)33 = 2.(9 - 1)33 = 2.[B(9) - 1] = B(9) - 2 = B(9) + 7 Vy: 2100 chia cho 9 th d 7 b) Tng t ta c:2100 = (210)10 = 102410 =[B(25) - 1]10=B(25) + 1 Vy: 2100 chia chop 25 th d 1 c)S dng cng thc Niutn: 2100 = (5 - 1)50 = (550 - 5. 549 + + 50.492. 52 - 50 . 5 ) + 1 Khng k phn h s ca khai trin Niutn th 48 s hng u cha tha s 5 vi s m ln hn hoc bng 3 nn u chia ht cho 53= 125, hai s hng tip theo: 50.492. 52 -50.5 cng chia ht cho 125 , s hng cui cng l 1 Vy: 2100 = B(125) + 1 nn chia cho 125 th d 1 Bi 2: Vit s 19951995 thnh tng ca cc s t nhin . Tng cc lp phng chia cho 6 th d bao nhiu? Gii t 19951995 = a = a1 + a2 + + an.

Gi 3 3 3 31 2 3 nS a a+ a+ ...+ a = += 3 3 3 31 2 3 na a+ a+ ...+ a ++ a - a = (a1 3 - a1) + (a2 3 - a2) + + (an 3 - an) + a Mi du ngoc u chia ht cho 6 v mi du ngoc l tch ca ba s t nhin lin tip. Ch cn tm s d khi chia a cho 6 1995 l s l chia ht cho 3, nn a cng l s l chia ht cho 3, do chia cho 6 d 3 Bi 3: Tm ba ch s tn cng ca 2100 vit trong h thp phn gii Tm 3 ch s tn cng l tm s d ca php chia 2100 cho 1000 Trc ht ta tm s d ca php chia 2100 cho 125 Vn dng bi 1 ta c 2100 = B(125) + 1 m 2100 l s chn nn 3 ch s tn cng ca n ch c thl 126, 376, 626 hoc 876 Hin nhin 2100 chia ht cho 8 v 2100 = 1625 chi ht cho 8 nn ba ch s tn cng ca n chia ht cho 8 trong cc s 126, 376, 626 hoc 876 ch c 376 chia ht cho 8 Vy: 2100 vit trong h thp phn c ba ch s tn cng l 376 Tng qut: Nu n l s chn khng chia ht cho 5 th 3 ch s tn cng ca n l 376 Bi 4: Tm s d trong php chia cc s sau cho 7 a) 2222 + 5555 b)31993 www.VNMATH.comwww.vnmath.com 11c) 19921993 + 19941995d)193023Gii a) ta c: 2222 + 5555 = (21 + 1)22 + (56 1)55 = (BS 7 +1)22 + (BS 7 1)55

= BS 7 + 1 + BS 7 - 1 = BS 7 nn2222 + 5555chia 7 d 0 b) Lu tha ca 3 st vi bi ca 7 l 33 = BS 7 1 Ta thy 1993 =BS 6 + 1 = 6k + 1, do : 31993 = 3 6k + 1 = 3.(33)2k = 3(BS 7 1)2k = 3(BS 7 + 1) = BS 7 + 3 c) Ta thy 1995 chia ht cho 7, do : 19921993 + 19941995 = (BS 7 3)1993 + (BS 7 1)1995 =BS 7 31993 + BS 7 1 Theo cu b ta c 31993 = BS 7 + 3 nn 19921993 + 19941995 = BS 7 (BS 7 + 3) 1 = BS 7 4 nn chia cho 7 th d 3d) 193023= 32860 = 33k + 1 = 3.33k = 3(BS 7 1) =BS 7 3 nn chia cho 7 th d 4Bi tp v nh Tms d khi: a) 21994 cho 7 b) 31998 + 51998 cho 13 c) A =13 + 23 + 33 + ...+ 993 chia cho B = 1 + 2 + 3 + ... + 99 Dng 3: Tm iu kin xy ra quan h chia ht Bi 1: Tmn e Z gi tr ca biu thc A = n3 + 2n2 - 3n + 2 chia ht cho gi tr ca biu thc B = n2 - n Gii Chia A cho B ta c: n3 + 2n2 - 3n + 2= (n + 3)(n2 - n) + 2 A chia ht cho B th 2 phi chia ht cho n2 - n = n(n - 1) do 2 chia ht cho n, ta c: n1- 12- 2 n - 10- 21- 3 n(n - 1)0226 loiloi Vy: gi tr ca biu thc A = n3 + 2n2 - 3n + 2 chia ht cho gi tr ca biu thcB = n2 - n th n{ } 1; 2 e Bi 2: a) Tm n e N n5 + 1 chia ht cho n3+ 1 b) Gii bi ton trn nu n e Z Gii Ta c:n5+ 1 n3 + 1 n2(n3 + 1) - (n2 - 1) n3 + 1 (n + 1)(n - 1) n3 + 1 (n + 1)(n - 1) (n + 1)(n2 - n + 1) n - 1 n2 - n + 1(V n + 1=0) a) Nu n = 1 th0 1 Nu n > 1 th n - 1 < n(n - 1) + 1 2 thn2 n + 2 khong la so chnh phng V (n 1)2 = n2 (2n 1) < n2 (n - 2) < n2 b) Ta con5 n chia het cho 5 V n5 n = (n2 1).n.(n2 + 1) Vi n = 5k th n chia het cho 5 Vi n = 5k1 th n2 1 chia het cho 5 Vi n = 5k2 th n2 + 1 chia het cho 5 Nenn5 n + 2 chia cho 5 th d 2 nenn5 n + 2 co ch so tan cung la 2 hoac 7 nen n5 n + 2 khong la so chnh phng Vay : Khong co gia tr nao cua nthoa man bai toan www.VNMATH.comwww.vnmath.com 21Bai 6 : a)Chng minh rang : Moi so le eu viet c di dang hieu cua hai so chnh phng b) Mot so chnh phng co ch so tan cung bang 9 th ch so hang chuc la ch so chan Giai Moi so le eu co dang a = 4k + 1 hoac a = 4k + 3 Vi a = 4k + 1 tha = 4k2 + 4k + 1 4k2 = (2k + 1)2 (2k)2 Vi a = 4k + 3 th a = (4k2 + 8k + 4) (4k2 + 4k + 1) = (2k + 2)2 (2k + 1)2 b)A la so chnh phng co ch so tan cung bang 9 nenA = (10k3)2 =100k260k + 9 = 10.(10k2 6) + 9 So chuc cua A la 10k26 la so chan (pcm) Bai 7:Mot so chnh phng co ch so hang chuc la ch so le. Tm ch so hang n v Giai Goi n2 = (10a + b)2 = 10.(10a2 + 2ab) + b2 nen ch so hang n v can tm la ch so tan cung cua b2 Theo e bai , ch so hang chuc cua n2 la ch so le nen ch so hang chuc cua b2 phai leXet cac gia tr cua b t 0 en 9 th ch co b2 = 16, b2 = 36 co ch so hang chuc la ch so le, chung eu tan cung bang 6 Vay : n2 co ch so hang n v la 6 * Bai tap ve nha: Bai 1:Cac so sau ay, so nao la so chnh phng a) A = 5022.....24b) B = 11115556c) C = n99....9 n00....025d) D = n44.....4 n - 188....89e) M =2n11.....1 n22....2 f) N = 12 + 22 + ...... + 562 Bai 2: Tm so t nhien n e cac bieu thc sau la so chnh phng a) n3 n + 2 b) n4 n+ 2 Bai 3: Chng minh rang a)Tong cua hai so chnh phng le khong la so chnh phng b) Mot so chnh phng co ch so tan cung bang 6 th ch so hang chuc la ch so le Bai 4: Mot so chnh phng co ch so hang chuc bang 5. Tm ch so hang n v www.vnmath.com www.VNMATH.comwww.vnmath.com 22 CHUYEN E 6 ONG D THC A. NH NGHA: Neu hai so nguyen a va b co cung so d trong phep chia cho mot so t nhien m=0 th ta noi a ong d vi b theo moun m, va co ong d thc: a b (mod m) V du:7 10 (mod 3) , 12 22 (mod 10) + Chu y: a b (mod m) a b m B. TNH CHAT: 1. Tnh chat phan xa: a a (mod m) 2. Tnh chat oi xng: a b (mod m) b a (mod m) 3. Tnh chat bac cau: a b (mod m), b c (mod m) th a c (mod m) 4. Cong , tr tng ve: ab (mod m)a c b d (mod m)cd (mod m) He qua: a) a b (mod m) a+ c b + c (mod m) b) a + b c (mod m) a c - b (mod m) c) a b (mod m) a + km b (mod m) 5. Nhan tng ve : ab (mod m)ac bd (mod m)cd (mod m) He qua: a) a b (mod m) ac bc(mod m) (c e Z) b) a b (mod m) an bn (mod m) 6. Co the nhan (chia) hai ve va moun cua mot ong d thc vi mot so nguyen dng a b (mod m) ac bc (mod mc) Chang han: 11 3 (mod 4) 22 6 (mod 8) 7. acbc (mod m)a b (mod m)(c, m) = 1 Chang han : 162 (mod 7)8 1 (mod 7)(2, 7) = 1 C. CAC V DU: 1. V du 1: Tm so d khi chia 9294 cho 15 Giai Ta thay92 2 (mod 15) 9294 294 (mod 15) (1) Lai co 24 1 (mod 15) (24)23. 22 4 (mod 15) hay 294 4 (mod 15) (2) T (1) va (2) suy ra9294 4 (mod 15) tc la 9294 chia 15 th d 4 2. V du 2: Chng minh: trong cac so co dang 2n 4(n e N), co vo so so chia het cho 5 www.VNMATH.comwww.vnmath.com 23That vay: T 24 1 (mod 5)24k 1 (mod 5) (1) Lai co 22 4 (mod 5) (2) Nhan (1) vi (2), ve theo ve ta co: 24k + 2 4 (mod 5) 24k + 2- 4 0 (mod 5) Hay 24k + 2- 4 chia het cho 5 vi moi k = 0, 1, 2, ... hay ta c vo so so dang 2n 4 (n e N) chia het cho 5 Chu y: khi giai cac bai toan ve ong d, ta thng quan tam en a1 (mod m) a 1 (mod m) an 1 (mod m) a -1 (mod m) an (-1)n (mod m) 3. V du 3: Chng minh rang a) 2015 1 chia het cho 11b) 230 + 330 chi het cho 13 c) 555222 + 222555 chia het cho 7 Giai a) 25 - 1 (mod 11) (1); 10 - 1 (mod 11) 105 - 1 (mod 11) (2) T (1) va (2) suy ra 25. 105 1 (mod 11) 205 1 (mod 11)205 1 0 (mod 11) b) 26 - 1 (mod 13)230 - 1 (mod 13) (3) 33 1 (mod 13) 3301 (mod 13) (4) T (3) va (4) suy ra 230 + 330 - 1 + 1 (mod 13) 230 + 330 0 (mod 13) Vay: 230 + 330 chi het cho 13 c) 555 2 (mod 7) 555222 2222 (mod 7) (5) 23 1 (mod 7) (23)74 1 (mod 7) 555222 1 (mod 7) (6) 222 - 2 (mod 7) 222555 (-2)555 (mod 7) Lai co (-2)3 - 1 (mod 7) [(-2)3]185 - 1 (mod 7) 222555 - 1 (mod 7) Ta suy ra 555222 + 222555 1 - 1 (mod 7) hay 555222 + 222555 chia het cho 7 4. V du 4:Chng minh rang so 4n + 122+ 7 chia het cho 11 vi moi so t nhien n That vay:Ta co: 25 - 1 (mod 11) 2101 (mod 11) Xet so d khi chia 24n + 1 cho 10. Ta co: 24 1 (mod 5) 24n 1 (mod 5) 2.24n 2 (mod 10) 24n + 1 2 (mod 10) 24n + 1 = 10 k + 2 Nen 4n + 122+ 7 = 210k + 2 + 7 =4. 210k + 7 = 4.(BS 11 + 1)k + 7 = 4.(BS 11 + 1k) + 7 = BS 11 + 11 chia het cho 11 Bai tap ve nha: Bai 1: CMR: a) 228 1 chia het cho 29 b)Trong cac so co dang2n 3 co vo so so chia het cho 13 Bai 2:Tm so d khi chia A =2011 + 2212 + 19962009 cho 7. www.vnmath.com www.VNMATH.comwww.vnmath.com 24CHUYEN E 7 CAC BAI TOAN VE BIEU THC HU T A. Nhac lai kien thc: Cac bc rut gon bieu thc hu t a) Tm KX: Phan tch mau thanh nhan t, cho tat ca cac nhan t khac 0 b) Phan tch t thanh nhan , chia t va mau cho nhan t chungB. Bai tap: Bai 1: Cho bieu thc A = 4 24 25 410 9x xx x + + a) Rut gon A b) tm x e A = 0 c) Tm gia tr cua A khi2 1 7 x =Giai a)kx :x4 10x2 + 9=0 [(x2)2 x2] (9x2 9)=0 x2(x2 1) 9(x2 1)=0 (x2 1)(x2 9)=0(x 1)(x + 1)(x 3)(x + 3)=0 13= = xx T : x4 5x2 + 4 = [(x2)2 x2] (x2 4) = x2(x2 1) 4(x2 1)=(x2 1)(x2 4)= (x 1)(x + 1)(x 2)(x + 2)Vi x= 1; x= 3 thA =(x - 1)(x + 1)(x - 2)(x + 2) (x - 2)(x + 2)(x - 1)(x + 1)(x - 3)(x + 3)(x - 3)(x + 3)=b) A = 0 (x - 2)(x + 2)(x - 3)(x + 3) = 0 (x 2)(x + 2) = 0 x =2 c)2 1 7 x = 2 1 7 2 8 42 1 7 2 6 3x x xx x x = = =

= = = * Vi x = 4 th A = (x - 2)(x + 2) (4 - 2)(4 + 2) 12(x - 3)(x + 3) (4 - 3)(4 + 3) 7= =* Vi x = - 3 th A khong xac nh 2. Bai 2: Cho bieu thcB = 3 23 22x 7x 12x 453x 19x 33x 9 + + a) Rut gon B b) Tm x e B > 0 Giaia) Phan tch mau: 3x3 19x2 + 33x 9 = (3x3 9x2) (10x2 30x) + (3x 9)= (x 3)(3x2 10x + 3) = (x 3)[(3x2 9x) (x 3)] = (x 3)2(3x 1) kx: (x 3)2(3x 1)=0 x=3 va x=13 b) Phan tch t, ta co: 2x3 7x2 12x + 45 = (2x3 6x2 )- (x2 - 3x) (15x - 45) = (x 3)(2x2 x 15) www.VNMATH.comwww.vnmath.com 25= (x 3)[(2x2 6x) + (5x 15)] = (x 3)2(2x + 5) Vi x=3 va x=13

ThB = 3 23 22 7 12 453 19 33 9x x xx x x + + = 22(x - 3) (2x + 5) 2x + 5(x - 3) (3x - 1) 3x - 1=c) B > 0 2x + 53x - 1 > 0 133 1 05 12 5 02 35 3 1 0 12 32 5 052xxx xxxx xxx >

> > >+ >

< < < + 0 th 2 x += x + 2 nenD = 3 2222 4x x xx x x+ + + = 3 2 222 ( 1)( 2)( 2) 4 ( 2) ( 2)( 2) 2x x x xx x x xxx x xx x x+ + = =+ + + + Neux + 2 < 0 th 2 x += - (x + 2) nen www.VNMATH.comwww.vnmath.com 26D = 3 2222 4x x xx x x+ + + = 3 222 ( 1)( 2)( 2) 4 ( 2) ( 2)( 2) 2x x x xx x xxx x xx x x+ + = = + + + + Neu x + 2 = 0 x = -2th bieu thc D khong xac nh b) e D co gia tr nguyen th22x x hoac 2x co gia tr nguyen +) 22x x co gia tr nguyen 2x(x - 1) 2x- x 2 x > - 2 x > - 2 V x(x 1) la tch cua hai so nguyen lien tiep nen chia het cho 2 vi moi x > - 2 +) 2x co gia tr nguyenx 2x = 2k 2k (k Z; k < - 1)x < - 2 x < - 2x = e c) Khia x = 6 x > - 2 nen D = 22x x = 6(6 1)152=* Dang 2: Cac bieu thc co tnh quy luat Bai 1: Rut gon cac bieu thc a) A = | |2 2 23 5 2 1......(1.2) (2.3)( 1)nn n++ + ++ Phng phap: Xuat phat t hang t cuoi e tm ra quy luat Ta co | |22 1( 1)nn n++ = 2 2 2 22 1 1 1( 1) ( 1)nn n n n+= + + Nen A = 2 2 2 2 2 2 2 2 2 21 1 1 1 1 1 1 1 1 1 ( 1)......1 2 2 3 3 ( 1) 1 ( 1) ( 1)n nn n n n n+ + + + = =+ + +

b) B = 2 2 2 21 1 1 11 . 1 . 1 ........ 12 3 4 n| | | | | | | | ||||\ . \ . \ . \ . Ta co22 2 21 1 ( 1)( 1)1k k kk k k + = =Nen B = 2 2 2 2 2 2 2 21.3 2.4 3.5 ( 1)( 1) 1.3.2.4...( 1)( 1) 1.2.3...( 1) 3.4.5...( 1) 1 1 1. . ... . .2 3 4 2 .3 .4 ... 2.3.4...( 1) 2.3.4.... 2 2n n n n n n n nn n n n n n n + + + + += = = = c) C = 150 150 150 150......5.8 8.11 11.14 47.50+ + + += 1 1 1 1 1 1 1150. . ......3 5 8 8 11 47 50| | + + + |\ . = 50.1 1 950. 455 50 10| | = = |\ . d) D = 1 1 1 1......1.2.3 2.3.4 3.4.5 ( 1) ( 1) n n n+ + + + + =1 1 1 1 1 1 1. ......2 1.2 2.3 2.3 3.4 ( 1) ( 1) n n n n| | + + + | +\ .= 1 1 1 ( 1)( 2)2 1.2 ( 1) 4 ( 1)n nn n n n ( + = (+ + Bai 2:a) Cho A = 1 2 2 1...1 2 2 1m mm n + + + + ; B = 1 1 1 1......2 3 4 n+ + + + . TnhAB Ta co www.VNMATH.comwww.vnmath.com 27A = 11 1 1 1... 1 1 ... 1 ... ( 1)1 2 2 1 1 2 2 1nn n n nn nn n n n| || | | |+ + + + + + + = + + + + ||| \ . \ .\ .

= 1 1 1 1 1 1 1... 1 ... nB1 2 2 1 2 2 1n nn n n n| | | |+ + + + + = + + + = || \ . \ . AB = n b) A = 1 1 1 1......1.(2n - 1) 3.(2n - 3) (2n - 3).3 (2n - 1).1+ + + +;B = 1 + 1 1......3 2n - 1+ +Tnh A : B Giai A = 1 1 1 1 1 1 11 ... 12n2n - 1 3 2n - 3 2n - 3 3 2n - 1 ( | | | | | | | |+ + + + + + + + ||||(\ . \ . \ . \ . 1 1 1 1 1 1 11 ...... ...... 12n3 2n - 1 2n - 3 2n - 1 2n - 3 31 1 1 1 1 A 1.2. 1 ...... .2.B 2n 3 2n - 1 2n - 3 2n B n ( | | | |= + + + + + + + + + ||(\ . \ . | |= + + + + = = |\ . Bai tap ve nha Rut gon cac bieu thc sau: a) 1 1 1+......+1.2 2.3 (n - 1)n+b) 2 2 2 22 2 2 21 3 5 n. . ......2 1 4 1 6 1 (n + 1) 1 c) 1 1 1+......+1.2.3 2.3.4 n(n + 1)(n +2)+* Dang 3: Rut gon; tnh gia tr bieu thc thoa man ieu kien cua bien Bai 1:Cho1x 3x+ = . Tnh gi tr ca cc biu thc sau : a) 221A xx= +; b) 331B xx= +; c) 441C xx= +;d) 551D xx= + . Li gii a) 2221 1A x x 2 9 2 7x x = + = + - = - = ; b) 3331 1 1B x x 3 x 27 9 18x x x = + = + - + = - = ; c) 24 24 21 1C x x 2 49 2 47x x = + = + - = - = ; d) 2 3 52 3 51 1 1 1A.B x x x x D 3x x x x = + + = + + + = + D = 7.18 3 = 123. Bai 2:Cho x y z + + = 2a b c (1); a b c+ + = 2x y z (2). Tnh gia tr bieu thcD = 22 2a b c + + x y z| || | | | |||\ . \ .\ . www.VNMATH.comwww.vnmath.com 28T (1)suy ra bcx + acy + abz = 0 (3) T (2) suy ra2 22 2 2 2a b c ab ac bc a b c ab ac bc + + + 2 . 4+ +4 2 .x y z xy xz yz x y z xy xz yz| | | | | | | || | | | | | | |+ + = = + + ||||||||\ . \ . \ . \ .\ . \ . \ . \ . (4) Thay (3) vao (4) ta coD = 4 2.0 = 4 Bai 3 a) Cho abc = 2; rut gon bieu thc A = a b 2cab + a + 2 bc + b + 1 ac + 2c + 2+ +Ta co :A = a ab 2c a ab 2cab + a + 2 abc + ab + a ac + 2c + 2 ab + a + 2 2 + ab + a ac + 2c + abc+ + = + += a ab 2c a ab 2 ab + a + 21ab + a + 2 2 + ab + a c(a + 2 + ab) ab + a + 2 2 + ab + a a + 2 + ab ab + a + 2+ + = + + = =b) Cho a + b + c = 0; rut gon bieu thc B = 2 2 22 2 2 2 2 2 2 2 2a b ca- b- c b- c- a c- b- a+ +T a + b + c = 0a = -(b + c) a2 = b2 + c2 + 2bc a2 - b2 - c2 = 2bc Tng t ta co: b2 - a2 - c2 = 2ac ; c2 - b2 - a2 = 2ab (Hoan v vong quanh), nen B = 2 2 2 3 3 3a b c a b c2bc 2ac 2ab 2abc+ ++ + =(1) a + b + c = 0 -a = (b + c) -a3 = b3 + c3 + 3bc(b + c) -a3 = b3 + c3 3abc a3 + b3 + c3 = 3abc (2) Thay (2) vao (1) ta co B = 3 3 3a b c 3abc 32abc 2abc 2+ += =(Vabc=0) c) Cho a, b, c tng oi mot khac nhau thoa man: (a + b + c)2 = a2 + b2 + c2 Rut gon bieu thcC = 2 2 22 2 2a b c + a+ 2bc b+ 2ac c+ 2ab+T (a + b + c)2 = a2 + b2 + c2 ab + ac + bc = 0 a2 + 2bc = a2 + 2bc (ab + ac + bc) = a2 ab + bc ac = (a b)(a c) Tng t: b2 + 2 ac=(b a)(b c) ; c2 + 2ab = (c a)(c b) C= 2 2 2 2 2 2a b c a b c + - (a - b)(a - c) (b - a)(b - c)(c - a)(c - b) (a - b)(a - c) (a - b)(b - c)(a - c)(b - c)+ = +=2 2 2a (b - c) b (a - c) c (b - c) (a - b)(a - c)(b - c) -1(a - b)(a - c)(b - c)(a - b)(a - c)(b - c) (a - b)(a - c)(b - c) (a - b)(a - c)(b - c)+ = =* Dang 4: Chng minh ang thc thoa man ieu kien cua bien 1. Bai 1: Cho 1 1 1 + + = 2a b c (1); 2 2 21 1 1+ + = 2a b c(2). Chng minh rang: a + b + c =abcT (1) suy ra2 2 2 2 2 21 1 1 1 1 1 1 1 1 1 1 1 + + + 2. + +4 2. + +4+ + a b c ab bc ac ab bc ac a b c| | | | | |= = |||\ . \ . \ . 1 1 1 a + b + c + +1 1ab bc ac abc= = a + b + c = abc www.VNMATH.comwww.vnmath.com 292. Bai 2: Cho a, b, c 0 v a + b + c 0 tha mn iu kin 1 1 1 1a b c a b c+ + =+ +. Chng minh rng trong ba s a, b, c c hai s i nhau.T suy ra rng :2009 2009 2009 2009 2009 20091 1 1 1a b c a b c+ + =+ +. Ta c : 1 1 1 1a b c a b c+ + =+ + 1 1 1 10a b c a b c+ + - =+ + a b a b0ab c(a b c)+ ++ =+ + a b 0 a bc(a b c) ab(a b). 0 (a + b)(b + c)(c + a) = 0 b c 0 b cabc(a b c)c a 0 c a + = = - + + + + = + = = - + + + = = -

T suy ra : 2009 2009 2009 2009 2009 2009 20091 1 1 1 1 1 1a b c a ( c) c a+ + = + + =-

2009 2009 2009 2009 2009 2009 20091 1 1a b c a ( c) c a= =+ + + - + 2009 2009 2009 2009 2009 20091 1 1 1a b c a b c+ + =+ +. 3. Bai 3:Cho a b c b c a + + b c a a b c + = +(1) chng minh rang : trong ba so a, b, c ton tai hai so bang nhau T (1) 2 2 2 2 2 2 2 2 2a c + ab+ bc= b c + ac+ a b a (b - c) - a(c b ) bc(c - b) = 0 + (c b)(a2 ac = ab + bc) = 0 (c b)(a b)( a c) = 0 pcm 4. Bai 4: Cho (a2 bc)(b abc) = (b2 ac)(a abc); abc=0 va a=b Chng minh rang: 1 1 1 + + = a + b + ca b c TGT a2b b2c- a3bc + ab2c2 = ab2 a2c ab3c + a2bc2 (a2b ab2) + (a2c b2c) = abc2(a b) + abc(a - b)(a + b) (a b)(ab + ac + bc) = abc(a b)(a + b + c) ab + ac + bc = a + b + c abc1 1 1+ + = a + b + ca b c 5. Bai 5:Cho a + b + c = x + y + z = a b c + + = 0x y z. Chng minh rang: ax2 + by2 + cz2 = 0 Tx + y + z = 0 x2 = (y + z)2 ; y2 = (x + z)2 ; z2 = (y + x)2 ax2 + by2 + cz2 = a(y + z)2 + b(x + z)2 + c (y + x)2 = = (b + c)x2 + (a + c)y2 + (a + b)z2 + 2(ayz + bxz + cxy) (1) Ta + b + c = 0 - a = b + c; - b = a + c; - c = a + b (2) Ta b c + + = 0x y z ayz + bxz + cxy = 0 (3). Thay (2), (3) vao (1); ta co:ax2 + by2 + cz2 = -( ax2 + by2 + cz2 ) ax2 + by2 + cz2 = 0 6. Bai 6:www.VNMATH.comwww.vnmath.com 30Cho a b c +0b - c c - a a - b+ = ;chng minh: 2 2 2a b c+0(b - c) (c - a) (a - b)+ =Ta b c +0b - c c - a a - b+ = 2 2a b c b ab + ac - c = b - c a - c b - a (a - b)(c - a)+ = 2 22a b ab + ac - c(b - c) (a - b)(c - a)(b - c)= (1)(Nhan hai ve vi1b - c) Tng t, ta co: 2 22b c bc + ba - a(c - a) (a - b)(c - a)(b - c)=(2) ; 2 22c a ac + cb - b(a - b) (a - b)(c - a)(b - c)=(3) Cong tng ve (1), (2) va (3) ta co pcm 7. Bai 7:Cho a + b + c = 0; chng minh: a - bb - c c - a c a b + + c a b a - b b - c c - a | || |+ + | |\ .\ . = 9 (1) ata - bb - c c - a = x ;; c a b y z = = c 1 a 1 b 1 =; a - bxb - c c - ay z= =(1)( )1 1 1x + y + z+ +9x y z| | = |\ . Ta co:( )1 1 1 y + z x + z x + yx + y + z+ +3+ + x y z x y z| | | |= + ||\ . \ . (2) Ta lai co: 2 2y + z b - c c - a c b bc + ac - a c c(a - b)(c - a - b) c(c - a - b). .xa b a - baba - bab(a - b) ab | |= + = = = |\ .

= | |2c 2c - (a + b + c)2cab ab=(3) Tng t, ta co:2x + z 2ay bc=(4) ; 2x + y 2bz ac=(5) Thay (3), (4) va (5) vao (2) ta co:( )1 1 1x + y + z+ +3x y z| | = |\ . + 2 2 2 2c2a2bab bc ac+ += 3 + 2 abc(a3 + b3 + c3 ) (6) Ta + b + c = 0 a3 + b3 + c3 = 3abc (7) ? Thay (7) vao (6) ta co:( )1 1 1x + y + z+ +3x y z| | = |\ . + 2 abc. 3abc = 3 + 6 = 9 Bai tap ve nha: Bai 1: Cho bieu thc A = 22 3 2: 13 2 5 6 1x x x xx x x x x | | | | + ||+ + + + \ . \ . a) Rut gon A b) Tm x e A = 0; A > 0 Bai 2: Cho bieu thc B = 3 23 23 7 5 12 4 3y y yy y y + + a) Rut gon B www.VNMATH.comwww.vnmath.com 31b) Tm so nguyen y e 2D2y + 3 co gia tr nguyen c) Tm so nguyen y e B> 1 Bai 3 : cho1 1 1 + +0x y z = ; tnh gia tr bieu thc A = 2 2 2yz xz xy+ + x y z HD: A = 3 3 3xyz xyz xyz + + x y z ; van dung a + b + c = 0 a3 + b3 + c3 = 3abc Bai 4: Cho a3 + b3 + c3 = 3abc ; Tnh gia tr bieu thc A = a b c+ 1+ 1+ 1b c a| || || | | | |\ .\ .\ . Bai 5: Cho x + y + z = 0; chng minh rang:3 0y z x z x yx y z+ + ++ + + =Bai 6: Cho a + b + c = a2 + b2 + c2 = 1; a b cx y z= = . Chng minh xy + yz + xz= 0 www.vnmath.com www.VNMATH.comwww.vnmath.com 32 CHUYEN E 8- CAC BAI TOAN VE NH L TA-LET A.Kien thc: 1. nh l Ta-let: * nh l Talet ABCMN // BCA `) AM AN = AB AC * He qua: MN // BC AM AN MN = AB AC BC=B. Bai tap ap dung: 1. Bai 1: Cho t giac ABCD, ng thang qua A song song vi BC cat BD E, ng thang qua B song song vi AD cat AC G a) chng minh: EG // CD b) Gia s AB // CD, chng minh rang AB2 = CD. EG Giai Goi O la giao iem cua AC va BD a) V AE // BC OE OA = OB OC (1) BG // AC OB OG = OD OA (2) Nhan (1) vi (2) ve theo ve ta co: OE OG = OD OC EG // CD b) Khi AB // CD th EG // AB // CD, BG // AD nen 2AB OA OD CD AB CD = =AB CD. EGEG OG OB AB EG AB= = =Bai 2:Cho ABC vuong tai A, Ve ra pha ngoai tam giac o cac tam giac ABD vuong can B, ACFvuong can C. Goi H la giao iem cua AB va CD, K la giao iem cua AC va BF. Chng minh rang: a) AH = AK b) AH2 = BH. CK Giaiat AB = c, AC = b.BD // AC (cung vuong goc vi AB)nenAH AC b AH b AH bHB BD c HB c HB + AHb + c= = = = Hay AH b AH b b.cAHABb + c cb + c b + c= = =(1) NMC BAHFKDCBAOG ED CBAwww.VNMATH.comwww.vnmath.com 33AB // CF (cung vuong goc vi AC) nen AK AB c AK c AK cKC CF b KC b KC + AKb + c= = = = Hay AK b AK c b.cAKACb + c bb + c b + c= = =(2) T (1) va (2) suy ra: AH = AK b) TAH AC bHB BD c= =va AK AB cKC CF b= =suy raAH KC AH KCHB AK HB AH= = (V AH = AK) AH2 = BH . KC 3. Bai 3:Chohnh bnh hanh ABCD, ng thang a i qua A lan lt cat BD, BC, DC theo th t tai E, K, G. Chng minh rang: a) AE2 = EK. EG b) 1 1 1AE AK AG= +c) Khi ng thang a thay oi v tr nhng van qua A th tch BK. DG co gia tr khong oi Giai a) V ABCD la hnh bnh hanh va K e BC nen AD // BK, theo he qua cua nh l Ta-let ta co: 2EK EB AE EK AE = =AE EK.EGAE ED EG AE EG = =b) Ta co: AE DE = AK DB ; AE BE = AG BDnen AE AE BE DE BD 1 1 =1 AE 1AK AG BD DB BD AK AG| |+ + = = + = |\ . 1 1 1AE AK AG= +(pcm) c) Ta co: BK AB BK a = = KC CG KC CG(1); KC CG KC CG = = AD DG b DG(2) Nhan (1) vi (2) ve theo ve ta co: BK a=BK. DG = abb DG khong oi (V a = AB; b = AD la o dai hai canh cua hnh bnh hanh ABCD khong oi) 4. Bai 4: Cho t giac ABCD, cac iem E, F, G, H theo th t chia trong cac canh AB, BC, CD, DA theo t so 1:2. Chng minh rang: a) EG = FH b) EG vuong goc vi FHGiai Goi M, N theo th t la trung iem cua CF, DG Ta co CM = 12 CF = 13BCBM 1= BC 3 BE BM 1= = BA BC 3 EM // AC EM BM 2 2=EM =ACAC BE 3 3= (1) Tng t, ta c: NF // BD NF CF 2 2 = NF =BDBD CB 3 3= (2) GbaEKDCB AQPONMH FGEDCBAwww.VNMATH.comwww.vnmath.com 34m AC = BD (3) T (1), (2), (3) suy ra : EM = NF (a) Tng t nh trn ta c: MG // BD, NH // AC v MG = NH = 13AC (b) Mt khc EM // AC; MG // BD V AC BDEM MG

0EMG = 90 (4) Tng t, ta c:

0FNH = 90 (5) T (4) v (5) suy ra 0EMG = FNH = 90(c) T (a), (b), (c) suy ra AEMG =AFNH (c.g.c) EG = FH b) Gi giao im ca EG v FH l O; ca EM v FH l P; caEM v FN l Q th

0PQF = 90 0QPF + QFP = 90m QPF = OPE(i nh), OEP = QFP( AEMG =AFNH) Suy ra 0EOP = PQF = 90 EO OP EG FH 5. Bi 5:Cho hnh thang ABCD c y nh CD. T D v ng thng song song vi BC, ct AC ti M v AB ti K, TC v ng thng song song vi AD, ct AB ti F, qua F ta li v ng thng song song vi AC, ct BC ti P. Chng minh rng a) MP // AB b) Ba ng thng MP, CF, DB ng quy Gii a) EP // AC CP AF = PB FB (1) AK // CD CM DC = AM AK (2) cc t gic AFCD, DCBK la cc hnh bnh hnh nnAF = DC, FB = AK (3) Kt hp (1), (2) v (3) ta cCP CMPB AM= MP // AB (nh lTa-lt o) (4) b) Gi I l giao im ca BD v CF, ta c: CP CMPB AM== DC DCAK FB= M DC DIFB IB=(Do FB // DC) CP DIPB IB= IP // DC // AB (5) T (4) v (5) suy ra : qua P c hai ng thng IP, PM cng song song vi AB // DC nn theo tin clt th ba im P, I, M thng hang hay MP i qua giao im ca CF v DB hay ba ng thngMP, CF, DB ng quy 6. Bi 6: ChoAABC c BC < BA. Qua C k ng thng vung goc vi tia phn gic BE ca

ABC; ng thng ny ct BE ti F v ct trung tuyn BD ti G. Chng minh rng on thng EG b on thngDF chia lm hai phn bng nhau Gii Gi K l giao im ca CF v AB; M l giao im ca DF v BC AKBC c BF va l phn gic va l ng cao nnAKBC cn ti B BK = BC v FC = FK IPF KMDCBAwww.VNMATH.comwww.vnmath.com 35Mt khc D l trung im AC nn DF l ng trung bnh ca AAKC DF // AK hay DM // AB Suy ra M l trung im ca BCDF = 12AK (DF l ng trung bnh ca AAKC), ta c BG BK = GD DF( do DF // BK) BG BK 2BK = GD DF AK=(1) Mt khc CE DC - DE DC AD1 1DE DE DE DE= = = (V AD = DC) CE AE - DE DC AD1 1DE DE DE DE= = = Hay CE AE - DE AE AB1 2 2DE DE DE DF= = = (v AEDE= ABDF: Do DF // AB) Suy ra CE AK + BK 2(AK + BK)2 2DE DE AK= = (Do DF = 12AK)CE 2(AK + BK) 2BK2DE AK AK= =(2) T (1) v (2) suy raBGGD = CEDE EG // BC Gi giao im ca EG v DF l O ta c OG OE FO= = MC MB FM| | |\ . OG = OE Bi tp v nh Bi 1: Cho t gic ABCD, AC v BD ct nhau ti O. ng thng qua O v song song vi BC ct AB E; ng thng song song vi CD qua O ct AD ti F a) Chng minhFE // BD b) T O k cc ng thng song song vi AB, AD ct BD, CD ti G v H.Chng minh: CG. DH = BG. CH Bi 2:Cho hnh bnh hnh ABCD, im M thuc cnh BC, im N thuc tia i ca tia BC sao cho BN = CM; cc ng thng DN, DM ct AB theo th t ti E, F. Chng minh: a) AE2 = EB. FE b) EB =2ANDF| | |\ .. EF MGKFD E CBAwww.VNMATH.comwww.vnmath.com 36CHUYEN E 9 CAC BAI TOAN S DUNG NH L TALET VA TNH CHAT NG PHAN GIAC A. Kien thc: 1. nh l Ta-let: * nh l Talet ABCMN // BCA `) AM AN = AB AC * He qua: MN // BC AM AN MN = AB AC BC=2. Tnh chat ng phan giac:AABC ,AD la phan giac goc A BD AB= CD AC ADla phan giac goc ngoai tai A: BD' AB= CD' AC B. Bai tap van dung 1. Bai 1: ChoAABC co BC = a, AB = b, AC = c, phan giac AD a) Tnh o dai BD, CD b) Tia phan giac BI cua goc B cat AD I; tnh t so: AIID Giai a) AD la phan giac cua

BAC nen BD AB cCD AC b= = BD c BD c acBD = CD + BD b + c a b + c b + c= = Do o CD = a - acb + c = abb + c b) BI la phan giac cua

ABC nen AI AB ac b + cc : ID BD b + c a= = =2. Bai 2: ChoAABC, co

B< 600 phan giac AD a) Chng minh AD < AB b) Goi AM la phan giac cuaAADC. Chng minh rang BC > 4 DM Giai a)Ta co

AADB = C + 2 > A + C2 =

00180 - B602=

ADB >

B AD < AB b) Goi BC = a, AC = b, AB = c, AD = d TrongAADC, AM la phan giac ta co DM AD = CM AC DM AD DM AD = = CM + DM AD + AC CD AD + ACD'CBAD C BAacbID C BAMD B CANMC BAwww.VNMATH.comwww.vnmath.com 37 DM = CD.AD CD. dAD + AC b + d= ; CD = abb + c( Van dung bai 1) DM = abd(b + c)(b + d) e c/mBC > 4 DM ta c/ma > 4abd(b + c)(b + d) hay (b + d)(b + c) > 4bd (1) That vay : do c > d (b + d)(b + c) > (b + d)2> 4bd . Bat ang thc (1) c c/m Bai 3: ChoAABC, trung tuyen AM, cac tia phan giac cua cac goc AMB , AMC cat AB, AC theo th t D va E a) Chng minh DE // BC b) Cho BC = a, AM = m. Tnh o dai DE c) Tm tap hp cac giao diem I cua AM va DE neuAABC co BC co nh, AM = m khong oi d)AABC co ieu kien g th DE la ng trung bnh cua no Giai a) MD la phan giac cua

AMB nen DA MBDB MA=(1) ME la phan giac cua

AMC nen EA MCEC MA=(2) T (1), (2) va gia thiet MB = MC ta suy ra DA EADB EC= DE // BC b) DE // BC DE AD AIBC AB AM= = . atDE = xxm - x 2a.m2x = a m a + 2m= c) Ta co: MI = 12 DE = a.ma + 2m khong oi I luon cach M mot oan khong oi nen tap hp cac iem I la ng tron tam M, ban knh MI = a.ma + 2m (Tr giao iem cua no vi BC d) DE la ng trung bnh cuaAABC DA = DB MA = MB AABC vuong A4. Bai 4:ChoAABC ( AB < AC) cac phan giac BD, CE a) ng thang qua D va song song vi BC cat AB K, chng minh E nam gia B va K b) Chng minh: CD > DE > BE Giai a) BD la phan giac nenAD AB AC AE AD AE = < = DC BC BC EB DC EB EBKB EB< EDMIC BAEDMKCBAwww.VNMATH.comwww.vnmath.com 38 E nam gia K va B b) Goi M la giao iem cua DE va CB. Ta co CBD = KDB(so le trong) KBD = KDB ma E nam gia K va B nen

KDB >

EDB

KBD >

EDB

EBD >

EDB EB < DE Ta lai co CBD + ECB = EDB + DEC

DEC>

ECB

DEC>

DCE (V

DCE =

ECB) Suy ra:CD > ED CD > ED > BE 5. Bai 5: ChoAABC . Ba ng phan giac AD, BE, CF. Chng minh a. DB EC FA. . 1DC EA FB =. b. 1 1 1 1 1 1AD BE CF BC CA AB+ + > + + .Giai a)AD la ng phan giac cua

BAC nen ta co: DB AB =DC AC (1) Tng t: vi cac phan giac BE, CF ta co: EC BC =EA BA (2) ; FA CA =FB CB (3) T (1); (2); (3) suy ra: DB EC FA AB BC CA. .= . .DC EA FB AC BA CB= 1 b) tAB = c , AC = b , BC = a , AD = da.Qua C k ng thng song song vi AD , ct tia BA H.Theo L Talt ta c: AD BACH BH= BA.CH c.CH cAD.CHBH BA + AH b + c= = =Do CH < AC + AH = 2bnn: 2abcdb c = + > + ||\ . \ . Chng minh tng t ta c :1 1 1 12bd a c| |> + |\ . V 1 1 1 12cd a b| |> + |\ . Nn:1 1 1 1 1 1 1 1 1 12a b cd d d b c a c a b (| | | | | |+ + > + + + + + |||(\ . \ . \ . 1 1 1 1 1 1 1.22a b cd d d a b c| | + + > + + |\ . 1 1 1 1 1 1a b cd d d a b c + + > + + ( pcm ) Bi tp v nh ChoAABC co BC = a, AC = b, AB = c (b > c), cac phan giac BD, CE a) Tnh o dai CD, BE roi suy ra CD > BE b) Ve hnh bnh hanh BEKD. Chng minh: CE > EK c) Chng minh CE > BD www.vnmath.com HFEDCBAwww.VNMATH.comwww.vnmath.com 39 CHUYEN E10 CAC BAI TOAN VE TAM GIAC ONG DANG A. Kien thc: * Tam giac ong dang: a) trng hp th nhat: (c.c.c) AABC ABC AB AC BC = = A'B' A'C' B'C' b) trng hp th nhat: (c.g.c) AABC ABC AB AC = A'B' A'C' ; A = A'c. Trng hp ong dang th ba (g.g) AABC ABC A = A' ; B = B'AH; AHla hai ng cao tng ng th: A'H'AH = k (T so ong dang); A'B'C'ABCSS = K2 B. Bai tap ap dung Bai 1: ChoAABC co B = 2 C, AB = 8 cm, BC = 10 cm.a)Tnh AC b)Neu ba canh cua tam giac tren la ba so t nhien lien tiep th moi canh la bao nhieu? Giai Cach 1: Tren tia oi cua tia BA lay iem E sao cho:BD = BC AACDAABC (g.g) AC ADAB AC=2AC AB. AD =AB.(AB + BD) == AB(AB + BC)= 8(10 + 8) = 144 AC = 12 cm Cach 2: Ve tia phan giac BE cua

ABC AABEAACB 2AB AE BE AE + BE AC =AC= AB(AB + CB) AC AB CB AB + CB AB + CB= = = = 8(8 + 10) = 144 AC = 12 cm b) Goi AC = b, AB = a, BC = c th t cau a ta co b2 = a(a + c) (1) V b > anen co the b = a + 1 hoac b = a + 2 + Neu b = a + 1 th (a + 1)2 = a2 + ac2a + 1 = aca(c 2) = 1 a = 1; b = 2; c = 3(loai) + Neu b = a + 2 tha(c 4) = 4 - Vi a = 1 th c = 8 (loai) - Vi a = 2 th c = 6 (loai) EDCBADC BAwww.VNMATH.comwww.vnmath.com 40- vi a = 4 th c = 6 ; b = 5 Vay a = 4; b = 5; c = 6 Bai 2: ChoAABC can tai A, ng phan giac BD; tnh BDbiet BC = 5 cm; AC = 20 cm Giai Ta co CD BC 1 = AD AC 4= CD = 4 cmva BC = 5 cm Bai toan tr ve bai 1Bai 3: ChoAABC can tai A va O la trung iem cua BC. Mot iem O di ong tren AB, lay iem E tren AC sao cho 2OBCE = BD. Chng minh rang a)ADBOAOCE b)ADOEADBOAOCE c) DO, EO lan lt la phan giac cua cac goc BDE, CED d) khoang cach t O en oan ED khong oi khi D di ong tren AB Giai a) T 2OBCE = BD CE OB = OB BD va B = C(gt)ADBOAOCE b) T cau a suy ra 2 3 O = E(1) VB, O ,C thang hang nen

03 O+ DOE EOC 180 + =(2) trong tam giac EOC th

02 E+ C EOC 180 + =(3) T (1), (2), (3) suy ra

DOE B C = = ADOE vaADBO co DO OE = DB OC (DoADBOAOCE)va DO OE = DB OB (Do OC = OB) va

DOE B C = = nenADOEADBOAOCE c) T cau b suy ra 1 2 D= D DO la phan giac cua cac goc BDE Cung t cau b suy ra 1 2 E= EEO la phan giac cua cac goc CED c) Goi OH, OI la khoang cach t O en DE, CE th OH = OI, ma O co nh nen OH khong oiOI khong oi khi D di ong tren AB Bai 4: (e HSG huyen Loc ha nam 2007 2008) ChoAABC can tai A, co BC = 2a, M la trung iem BC, lay D, E thuoc AB, AC sao cho DME = Ba) Chng minh tch BD. CE khong oi b)Chng minh DM la tia phan giac cua

BDEc) Tnh chu vi cuaAAED neuA ABC la tam giac eu Giai 21321HIOEDC BAwww.VNMATH.comwww.vnmath.com 41a) Ta co

DMC = DME + CME = B + BDM, ma DME = B(gt) nen CME = BDM, ket hp vi B = C ( AABCcan tai A) suy raABDMACME (g.g) 2BD BM = BD. CE = BM. CM = aCM CE khong oi b)ABDMACME DM BD DM BD = = ME CM ME BM(do BM = CM) ADMEADBM (c.g.c) MDE = BMD hay DM la tia phan giac cua

BDEc) chng minh tng t ta co EM la tia phan giac cua

DEC keMHCE ,MIDE, MKDB th MH = MI = MK ADKM =ADIMDK =DIAEIM =AEHMEI = EH Chu viAAED la PAED = AD + DE + EA = AK +AH = 2AH (V AH = AK) A ABC la tam giac eu nen suy raA CME cung la tam giac euCH = MC2 2a= AH = 1,5a PAED = 2 AH = 2. 1,5 a = 3a Bai 5:Cho tam giac ABC, trung tuyen AM. Qua iem D thuoc canh BC, ve ng thang song song vi AM, cat AB, AC tai E va F a) chng minh DE + DF khong oi khi D di ong tren BC b) Qua A ve ng thang song song vi BC, cat FE tai K. Chng minh rang K la trung iem cua FE Giai a) DE // AM DE BD BD =DE =.AMAM BM BM(1) DF // AM DF CD CD CD =DF =.AM=.AM AM CM CM BM(2) T (1) va (2) suy raDE + DF = BD CD.AM +.AMBM BM = BD CD BC+.AM =.AM = 2AMBM BM BM| | |\ . khong oi b) AK // BC suy raAFKAAAMC (g.g) FK KA = AM CM (3) EK KA EK KA EK KA EK KA EK KA = = = ED BD ED + EK BD + KA KD BD + DM AM BM AM CM = =(2) (V CM = BM) T (1) va (2) suy raFK EKAM AM= FK = EK hay K la trung iem cua FE Bai 6: (e HSG huyen Thach ha nam 2003 2004) Cho hnh thoi ABCD canh a co

0A = 60 , mot ng thang bat ky qua C cat tia oi cua cac tia BA, DA tai M, N KHIMEDC BAKFEDMCBAwww.VNMATH.comwww.vnmath.com 42a) Chng minh rang tch BM. DN co gia tr khong oi b) Goi K la giao iem cua BN va DM. Tnh so o cua goc BKD Giai a) BC // AN MB CM = BA CN(1) CD// AM CM AD = CN DN (2) T (1) va (2) suy ra 2MB AD =MB.DN = BA.AD = a.a = aBA DN b)AMBD va ABDN co

MBD = BDN= 1200 MB MB CM AD BD = = BD BA CN DN DN= = (Do ABCD la hnh thoi co

0A = 60 nen AB = BC = CD = DA) AMBDABDN Suy ra 1 1 M= B .AMBD va ABKD co BDM = BDK va 1 1 M= Bnen 0BKD = MBD= 120Bai 7:Cho hnh bnh hanh ABCD co ng cheo ln AC,tia Dx cat SC, AB, BC lan lt tai I, M, N. Ve CE vuong goc vi AB, CF vuong goc vi AD, BG vuong goc vi AC. Goi K la iem oi xng vi D qua I. Chng minh rang a) IM. IN = ID2 b) KM DM = KN DN c) AB. AE + AD. AF = AC2 Giai a) T AD // CM IM CI = ID AI (1) T CD // AN CI ID AI IN= (2) T (1) va (2) suy ra IMID= IDIN hay ID2 = IM. IN b) Ta co DM CM DM CM DM CM = = = MN MB MN + DM MB + CM DN CB (3) TID = IK va ID2 = IM. INsuy ra IK2 = IM. IN IK IN IK - IM IN - IK KM KN KM IM = = = = IM IK IM IK IM IK KN IK KM IM CM CM = KN ID AD CB= =(4) T (3) va (4) suy ra KM DM = KN DN c) Ta co AAGBAAECAE AC=AB.AE = AC.AGAG AB AB. AE = AG(AG + CG) (5) ACGBAAFCAF CG CG = AC CB AD= (v CB = AD)AF . AD = AC. CG AF . AD = (AG + CG) .CG (6) 11KMNDCBAIKFGEMDCBANwww.VNMATH.comwww.vnmath.com 43Cong (5) va (6) ve theo ve ta co:AB. AE + AF. AD = (AG + CG) .AG + (AG + CG) .CG AB. AE + AF. AD = AG2 +2.AG.CG + CG2 = (AG + CG)2 = AC2 Vay:AB. AE + AD. AF = AC2 Bai tap ve nha Bai 1 Cho Hnh bnh hanh ABCD, mot ng thang cat AB, AD, AC lan lt tai E, F, G Chng minh: AB AD AC + = AE AF AG HD: Ke DM // FE, BN // FE (M, N thuoc AC) Bai 2: Qua nh C cua hnh bnh hanh ABCD, ke ng thang cat BD, AB, AD E, G, F chng minh: a) DE2 = FEEG. BE2 b) CE2 = FE. GE (Gi y: Xet cac tam giac DFE va BCE, DEC va BEG) Bai 3 Cho tam giac ABC vuong tai A, ng cao AH, trung tuyen BM, phan giac CD cat nhau tai mot iem. Chng minh rang a) BH CM AD. . 1HC MA BD =b) BH = AC www.VNMATH.comwww.vnmath.com 44 CHUYEN E 11 PHNG TRNH BAC CAO A.Muc tieu: * Cung co, on tap kien thc va ky nang giai cac Pt bac cao bang cach phan tch thanh nhan t * Khac sau ky nang phan tch a thc thanh nhan t va ky nang giai Pt B. Kien thc va bai tap: I. Phng phap: * Cach 1: e giai cac Pt bac cao, ta bien oi, rut gon e da Pt ve dang Pt co ve trai la mot a thc bac cao, ve phai bang 0, van dung cac phng phap phan tch a thc thanh nhan t e a Pt ve dang pt tch e giai * Cach 2: at an phu II. Cac v du: 1.V du 1: Giai Pt a) (x + 1)2(x + 2) + (x 1)2(x 2) = 12 ... 2x3 + 10x= 12x3 + 5x 6 = 0(x3 1) + (5x 5)(x 1)(x2 + x + 6) = 0 22x = 1x - 1 = 0x11 23x+ x + 6 = 0 x +02 4

= | |

+ =|

\ . (V21 23x +02 4| | + = |\ . vo nghiem) b) x4 + x2 + 6x 8 = 0 (1) Ve phai cua Pt la mot a thc co tong cac he so bang 0, nen co mot nghiem x = 1 nen co nhan t la x 1, ta co (1) (x4 x3) + (x3 x2) + (2x2 2x) + (8x 8) = 0 ... (x 1)(x3 + x2 + 2x + 8)(x 1)[(x3 + 2x2) (x2 + 2x) + (4x 8) ] = 0 (x 1)[x2(x + 2) x(x + 2) + 4(x + 2) = 0(x 1)(x + 2)(x2 x + 4) = 0 .... c) (x 1)3 + (2x + 3)3 = 27x3 + 8 x3 3x2 + 3x 1 + 8x3 + 36x2 + 54x + 27 27x3 8 = 0 - 18x3 + 33x2 + 57 x + 18 = 0 6x3 - 11x2 - 19x - 6= 0 (2) Ta thay Pt co mot nghiem x = 3, nen ve trai co nhan t x 3: (2)(6x3 18x2) + (7x2 21x) + (2x 6) = 0 6x2(x 3) + 7x(x 3) + 2(x 3) = 0 (x 3)(6x2 + 7x + 2) = 0 (x 3)[(6x2 + 3x) + (4x + 2)] = 0 (x 3)[3x(2x + 1) + 2(2x + 1)] = 0 (x 3)(2x + 1)(3x + 2) ..... d)(x2 + 5x)2 2(x2 + 5x) = 24[(x2 + 5x)2 2(x2 + 5x) + 1] 25 = 0 (x2 + 5x - 1)2 25 = 0(x2 + 5x - 1 + 5)( (x2 + 5x - 1 5) = 0 (x2 + 5x + 4) (x2 + 5x 6) = 0[(x2 + x) +(4x + 4)][(x2 x) + (6x 6)] = 0 (x + 1)(x + 4)(x 1)(x + 6) = 0 .... e)(x2 + x + 1)2 = 3(x4 + x2 + 1) (x2 + x + 1)2 - 3(x4 + x2 + 1) = 0 www.VNMATH.comwww.vnmath.com 45 (x2 + x + 1)2 3(x2 + x + 1)( x2 - x + 1) = 0 ( x2 + x + 1)[ x2 + x + 1 3(x2 - x + 1)] = 0( x2 + x + 1)( -2x2 + 4x - 2) = 0 (x2 + x + 1)(x2 2x + 1) = 0( x2 + x + 1)(x 1)2 = 0... f)x5 = x4 + x3 + x2 + x + 2 (x5 1) (x4 + x3 + x2 + x + 1) = 0 (x 1) (x4 + x3 + x2 + x + 1) (x4 + x3 + x2 + x + 1) = 0 (x 2) (x4 + x3 + x2 + x + 1) = 0+) x 2 = 0 x = 2 +) x4 + x3 + x2 + x + 1 = 0 (x4 + x3) + (x + 1) + x2 = 0(x + 1)(x3 + 1) + x2 = 0 (x + 1)2(x2 x + 1) + x2 = 0 (x + 1)2 [(x2 2.x.12 + 14) + 34] + x2 = 0 (x + 1)2 21 3x ++ 2 4 (| | ( |\ . ( + x2 = 0 Vo nghiem v(x + 1)2 21 3x ++ 2 4 (| | ( |\ . ( > 0 nhng khong xay ra dau bang Bai 2: a) (x2 + x - 2)( x2 + x 3) = 12 (x2 + x 2)[( x2 + x 2) 1] 12 = 0 (x2 + x 2)2 (x2 + x 2) 12 = 0at x2 + x 2 = y Th (x2 + x 2)2 (x2 + x 2) 12 = 0 y2 y 12 = 0(y 4)(y + 3) = 0 * y 4 = 0 x2 + x 2 4 = 0 x2 + x 6 = 0(x2 + 3x) (2x + 6) = 0 (x + 3)(x 2) = 0.... * y + 3 = 0 x2 + x 2 + 3 = 0 x2 + x + 1 = 0 (vo nghiem) b) (x 4)( x 5)( x 6)( x 7) = 1680(x2 11x + 28)( x2 11x + 30) = 1680 at x2 11x + 29 = y , ta co: (x2 11x + 28)( x2 11x + 30) = 1680(y + 1)(y 1) = 1680y2 = 1681y =41 y = 41 x2 11x + 29= 41 x2 11x 12 = 0 (x2 x) + (12x 12) = 0 (x 1)(x + 12) = 0..... * y = - 41 x2 11x + 29 = - 41 x2 11x + 70 = 0(x2 2x. 112+1214)+1594 = 0 c) (x2 6x + 9)2 15(x2 6x + 10) = 1 (3) at x2 6x + 9 = (x 3)2 = y> 0, ta co (3)y2 15(y + 1) 1 = 0y2 15y 16= 0 (y + 1)(y 15) = 0Vi y + 1 = 0 y = -1 (loai) Vi y 15 = 0y = 15 (x 3)2 = 16 x 3 =4 + x 3 = 4 x = 7 + x 3 = - 4x = - 1 d) (x2 + 1)2 + 3x(x2 + 1) + 2x2 = 0 (4) atx2 + 1 = y th (4) y2 + 3xy + 2x2 = 0(y2 + xy) + (2xy + 2x2) = 0(y + x)(y + 2x) = 0 www.VNMATH.comwww.vnmath.com 46+) x + y = 0x2 + x + 1 = 0 : Vo nghiem +) y + 2x = 0 x2 + 2x + 1 = 0 (x + 1)2 = 0 x = - 1 Bai 3: a) (2x + 1)(x + 1)2(2x + 3) = 18 (2x + 1)(2x + 2)2(2x + 3) = 72. (1) at 2x + 2 = y, ta co (1) (y 1)y2(y + 1) = 72y2(y2 1) = 72 y4 y2 72 = 0aty2 = z > 0Thy4 y2 72 = 0 z2 z 72 = 0 (z + 8)( z 9) = 0 * z + 8 = 0 z = - 8 (loai) * z 9 = 0 z = 9y2 = 9 y =3x = ... b)(x + 1)4 + (x 3)4 = 82 (2) at y = x 1x + 1 = y + 2; x 3 = y 2, ta co (2)(y + 2)4 + (y 2)4 = 82y4 +8y3 + 24y2 + 32y + 16 + y4 - 8y3 + 24y2 - 32y + 16 = 82 2y4 + 48y2 + 32 82 = 0 y4 + 24y2 25 = 0 at y2 = z> 0 y4 + 24y2 25 = 0 z2 + 24 z 25 = 0 (z 1)(z + 25) = 0 +) z 1 = 0 z = 1y = 1x = 0; x = 2 +) z + 25 = 0z = - 25 (loai) Chu y: Khi giai Pt bac 4 dang (x + a)4 + (x + b)4 = c ta thng at an phu y = x + a + b2c) (4 x)5 + (x 2)5 = 32(x 2)5 (x 4)5 = 32 at y = x 3x 2 = y + 1; x 4 = y 1; ta co: (x 2)5 (x 4)5 = 32 (y + 1)5 -(y 1)5 = 32 y5 + 5y4 + 10y3 + 10y2 + 5y + 1 (y5 - 5y4 + 10y3 - 10y2 + 5y - 1) 32 = 0 10y4 + 20y2 30 = 0 y4 + 2y2 3 = 0 at y2 = z > 0 y4 + 2y2 3 = 0z2 + 2z 3 = 0(z 1)(z + 3) = 0 ........ d) (x - 7)4 + (x 8)4 = (15 2x)4

at x 7 = a; x 8 = b ; 15 2x = c th- c = 2x 15 a + b = - c , Nen (x - 7)4 + (x 8)4 = (15 2x)4a4 + b4 = c4 a4 + b4 - c4 = 0 a4 + b4 (a + b)4 = 0 4ab(a2 + 32ab + b2) = 0 223 74ab a +b+ b4 16 (| | ( |\ . ( = 0 4ab = 0(V 223 7a +b+ b4 16| | |\ .> 0 nhng khong xay ra dau bang) ab = 0x = 7; x = 8 e) 6x4 + 7x3 36x2 7x + 6 = 0221 16 x 7 x -36 0x x| | | |+ + = ||\ . \ .

(V x = 0 khong la nghiem). at 1x - x = y 221xx+= y2 + 2 , th www.VNMATH.comwww.vnmath.com 47221 16x 7 x -36 0x x| | | |+ + = ||\ . \ . 6(y2 + 2) + 7y 36 = 06y2 + 7y 24 = 0 (6y2 9y) + (16y 24) = 0 (3y + 8 )(2y 3)= 0 +) 3y + 8 = 0y = - 83 1x - x = - 83... (x + 3)(3x 1) = 0x = - 3x + 3 = 013x - 1 = 0 x = 3

+) 2y 3 = 0y =32 1x - x = 32 ... (2x + 1)(x 2) = 0x = 2x - 2 = 012x + 1 = 0 x = - 2

Bai 4: Chng minh rang: cac Pt sau vo nghiem a) x4 3x2 + 6x + 13 = 0( x4 4x2+ 4) +(x2 + 6x + 9) = 0 (x2 2)2 + (x + 3)2 = 0 Ve trai (x2 2)2 + (x + 3)2> 0 nhng khong ong thi xay ra x2 = 2 va x = -3 b) x6 + x5 + x4 + x3 + x2 + x + 1 = 0 (x 1)( x6 + x5 + x4 + x3 + x2 + x + 1) = 0 x7 1 = 0 x = 1 x = 1 khong la nghiem cua Ptx6 + x5 + x4 + x3 + x2 + x + 1 = 0 Bai tap ve nha: Bai 1: Giai cac Pt a)(x2 + 1)2 = 4(2x 1) HD: Chuyen ve, trien khai (x2 + 1)2, phan tch thanh nhan t: (x 1)2(x2 + 2x + 5) = 0 b) x(x + 1)(x + 2)(x + 3) = 24(Nhan 2 nhan t vi nhau, ap dung PP at an phu) c)(12x + 7)2(3x + 2)(2x + 1) = 3 (Nhan 2 ve vi 24, at 12x + 7 = y) d) (x2 9)2 = 12x + 1 (Them, bt 36x2) e) (x 1)4 + (x 2)4 = 1 ( at y = x 1,5; s: x = 1; x = 2) f) (x 1)5 + (x + 3)5 = 242(x + 1)(at x + 1 = y; s:0; -1; -2 ) g) (x + 1)3 + (x - 2)3 = (2x 1)3 at x + 1 = a; x 2 = b; 1 - 2x= cth a + b + c = 0a3 + b3 + c3 = 3abc h) 6x4 + 5x3 38x2 + 5x + 6 = 0(Chia 2 ve cho x2; aty = 1x + x ) i) x5 + 2x4 + 3x3 + 3x2 + 2x + 1 = 0(Ve trai la a thc co tong cac he so bac chan bang tong cac he so bac le...) Bai 2: Chng minh cac pt sau vo nghiem a) 2x4 10x2 + 17 = 0 (Phan tch ve trai thanh tong cua hai bnh phng) b) x4 2x3 + 4x2 3x + 2 = 0(Phan tch ve trai thanh tch cua 2 a thc co gia tr khong am....) www.VNMATH.comwww.vnmath.com 48 CHUYEN E 12 VE NG THANG SONG SONG E TAO THANH CAC CAP OAN THANG TY LE A. Phng phap: Trong cac bai tap van dung nh l Talet. Nhieu khi ta can ve them ng phla mot ng thang song song vi mot ng thang cho trc,. ay la mot cach ve ng phu hay dung, v nh o ma tao thanh c cac cap oan thang t le B. Cac v du: 1) V du 1: Tren cac canh BC, CA, ABcua tam giac ABC, lay tng ng cac iem P, Q, Rsao cho ba ng thang AP, BQ, CR cat nhau tai mot iem.Chng minh: AR BP CQ. . 1RB PC QA =(nh l Ce va) Giai Qua A ke ng thang song song vi BC cat cac ng thang CR, BQ tai E, F. Goi O la giao iem cua AP, BQ, CR AAREABRC AR AE = RB BC (a) ABOPAFOA BP OP = FA OA (1) APOCAAOE PC PO = AE AO =(2) T (1) va (2) suy ra: BP PC BP FA = FA AE PC AE =(b) AAQFACQB CQ BC = AQ FA (c) Nhan (a), (b), (c) ve theo ve ta co: AR BP CQ AE FA BC. . . . 1RB PC QA BC AE FA= =* ao lai: NeuAR BP CQ. . 1RB PC QA = th bai ng thang AP, BQ, CR ong quy 2) V du 2:Mot ng thang bat ky cat cac canh( phan keo dai cua cac canh) cua tam giac ABC taiP, Q, R.Chng minh rang: RB.QA.PC1RA.CQ.BP = (nh l Me-ne-la-uyt) Giai: Qua A ke ng thang song song vi BC cat PR tai E. Ta co ARAEARBP RB BP = RA AE (a) OFERQC P BAERQC P BAwww.VNMATH.comwww.vnmath.com 49AAQEACQP QA AE = QC CP (b) Nhan ve theo ve cac ang thc (a) va (b) ta coRB QA BP AE. =.RA QC AE CP (1) Nhan hai ve ang thc (1) vi PCBPta co: RB PC QA BP AE PC. .=. . 1RA BP QC AE CP BP =ao lai: NeuRB.QA.PC1RA.CQ.BP =th ba iem P, Q, R thang hang 3) V du 3: Cho tam giac ABC, trung tuyen AM. Goi I la iem bat ky tren canh BC. ng thang qua I song song vi AC cat AB K; ng thang qua I song song vi AB cat AC, AM theo th t D, E. Chng minh DE = BK Giai Qua M ke MN // IE (Ne AC).Ta co: DE AE DE MN = MN AN AE AN =(1) MN // IE, ma MB = MC AN = CN (2) T (1) va (2) suy raDE MNAE CN=(3) Ta lai co MN CN MN ABAB AC CN AC= = (4) T (4) va (5) suy ra DE ABAE AC=(a) Tng t ta co: BK ABKI AC=(6) V KI // AC, IE // AC nen t giac AKIE la hnh bnh hanh nen KI = AE (7) T (6) va (7) suy ra BK BK ABKI AE AC= =(b) T (a) va (b) suy ra DE BKAE AE= DE = BK 4) V du 4: ng thang qua trung iem cua canh oi AB, CD cua t giac ABCD cat cac ng thang AD, BC theo th t I, K. Chng minh: IA . KC = ID. KB Giai Goi M, N theo th t la trung iem cua AB, CD Ta co AM = BM; DN = CN Ve AE, BF lan lt song song vi CD AAME =ABMF (g.c.g) AE = BFNDIMEKCBAFEIKMN DCBAwww.VNMATH.comwww.vnmath.com 50 Theo nh l Talet ta co:IA AE BF = ID DN CN=(1) Cung theo nh l Talet ta co: KB BF = KC CN(2) T (1) va (2) suy ra IA KB =ID KC IA . KC = ID. KB 5) V du 5:Cho

xOy , cac iem A, B theo th t chuyen ong tren cac tia Ox, Oy sao cho 1 1 1 + OA OB k=(k la hang so). Chng minh rang AB luon i qua mot iem co nh Giai Ve tia phan giac Ozcua

xOycat AB C. ve CD // OA (D e OB)

DOC = DCO = AOC ACOD can tai D DO = DC Theo nh l Talet ta co CD BD CD OB - CD = OA OB OA OB = CD CD 1 1 11OA OB OA OB CD+ = + =(1) Theo gia thiet th1 1 1 + OA OB k=(2) T (1) va (2) suy ra CD = k , khong oiVay AB luon i qua mot iem co nh la C sao cho CD = k va CD // Ox , D e OB 6) V du 6: Cho iem M di ong tren ay nho AB cua hnh thang ABCD, Goi O la giao iem cua hai canh ben DA, CB. Goi G la giao iem cua OA va CM, H la giao iem cua OB va DM. Chng minh rang: Khi M di ong tren AB th tong OG OH + GD HC khong oi Giai Qua O ke ng thang song vi AB cat CM, DM theo th t I va K. Theo nh l Talet ta co: OG OIGD CD= ; OH OKHC CD= OG OH OI OK IK + GD HC CD CD CD= + = OG OH IK + GD HC CD = (1) Qua M ve ng thang vuong goc vi AB cat IK, CD theo th t P va Q, ta co: IK MP FOCD MQ MQ= =khong oi v FO la khoang cach t O en AB, MQ la ng cao cua hnh thang nen khong oi (2) T (1) va (2) suy raOG OH FO + GD HC MQ=khong oi QPFK IHGMOD CBAzOyxDCBAwww.VNMATH.comwww.vnmath.com 51 www.vnmath.com 7) V du 7: Cho tam giac ABC (AB < AC), phan giac AD. Tren AB lay iem M, tren AC lay iem N sao cho BM = CN, goi giao iem cua CM va BN la O, T O ve ng thang song song vi AD cat AC, AB tai E va F. Chng minh rang: AB = CF;BE = CA Giai.AD la phan giac nen BAD = DAF EI // AD BAD = AEF(goc ong v) Ma DAF OFC =(ong v); AFE = OFC(oi nh) Suy ra AEF AFE = AAFE can tai A AE =AF (a) Ap dung nh l Talet vaoAACD , vi I la giao iem cua EF vi BC ta co CF CI CF CA = CA CD CI CD =(1) AD la phan giac cua

BAC nen CA BACD BD=(2) T (1) va (2) suy ra CF BACI BD=(3) Ke ng cao AG cuaAAFE . BP // AG (P eAD); CQ // AG (Qe OI) th BPD = CQI= 900 Goi trung iem cua BC la K, ta coABPK =ACQK (g.c.g) CQ = BP ABPD =ACQI (g.c.g) CI = BD (4) Thay (4) vao (3) ta co CF BABD BD= CF = BA (b) T (a) va (b) suy ra BE = CA Bai tap ve nha 1) Cho tam giac ABC. iem D chia trong BC theo t so 1 : 2, iem O chia trong AD theo t so 3 : 2. goi K la giao iem cua BO va AC. Chng minh rang KAKC khong oi 2) Cho tam giac ABC (AB > AC). Lay cac iem D, E tuy y th t thuoc cac canh AB, AC sao cho BD = CE. Goi giao iem cua DE, BC la K, chng minh rang : T so KEKD khong oi khi D, E thay oi tren AB, AC (HD: Ve DG // EC (G e BC). GPOKINDQC BMAFEwww.VNMATH.comwww.vnmath.com 52 CHUYEN E 13 BO E HNH THANG VA CHUM NG THANG ONG QUY A. Kien thc: 1) Bo e hnh thang: Trong hnh thang co hai ay khong bang nhau, ng thang i qua giao iem cua cac ng cheo va i qua giao iem cua cac ng thang cha hai canh ben th i qua trung iem cua hai ay Chng minh: Goi giao iem cua AB, CD la H, cua AC, BD la G, trung iem cua AD, BC la E va FNoi EG, FG, ta co:AADGACBG (g.g) , nen : AD AG 2AE AG AE AGCB CG 2CF CG CF CG= = =(1) Ta lai co : EAG FCG =(SL trong )(2) T (1) va (2) suy ra :AAEGACFG (c.g.c) Do o: AGE CGF = E , G , H thang hang (3) Tng t, ta co:AAEHABFH AHE BHF = H , E , Fthang hang (4) T (3) va (4) suy ra : H , E , G , Fthang hang 2) Chum ng thang ong quy: Neu cac ng thang ong quy cat hai ng thang song song th chung nh ra tren hai ng thang song song ay cac oan thang tng ng t le Neu m // n, ba ng thang a, b, c ong quy O chung cat m tai A, B, C va cat n tai A, B, C thAB BC AC = A'B' B'C' A'C'= hoac AB A'B' AB A'B' =; BC B'C' AC A'C'=* ao lai: + Neu ba ng thang trong o co hai ng thang cat nhau, nh ra tren hai ng thang song song cac cap oan thang tng ng t le th ba ng thang o ong quy + Neu hai ng thang b cat bi ba ng thang ong quy tao thanh cac cap oan thang tng ng t le th chung song song vi nhau B. Ap dung: 1) Bai 1: Cho t giac ABCD co M la trung iem CD, N la trung iem CB. Biet AM, AN cat BD thanh ba oan bang nhau. Chng minh rang ABCD la hnh bnh hanh Giai // /// /HGEFDC BAc baOnmA'B'C'CBAwww.VNMATH.comwww.vnmath.com 53Goi E, F la giao iem cua AM, AN vi BD; G, H la giao iem cua MN vi AD, BD MN // BC (MN la ng trung bnh cuaABCD) T giac HBFM la hnh thang co hai canh ben ong quy tai A, N la trung iem cua ay BF nen theo bo e hnh thang th N la trung iem cua ay MHMN = NH (1) Tng t : trong hnh thang CDEN th M la trung iem cua GN GM = MN (2) T (1) va (2) suy ra GM = MN = NH Ta coABNH =ACNM (c.g.c) BHN = CMN BH // CM hay AB // CD (a) Tng t:AGDM =ANCM (c.g.c) DGM = CNM GD // CN hay AD // CB (b) T (a) va (b) suy ra t giac ABCD co cac cap canh oi song song nen la hnh bnh hanh 2) Bai 2: ChoAABC co ba goc nhon, trc tam H, mot ng thang qua H cat AB, AC th t ta P, Q sao cho HP = HQ. Goi M la trung iem cua BC. Chng minh: HM PQ Giai Goi giao iem cua AH va BC la I T C ke CN // PQ (Ne AB),ta chng minh MHCN HMPQ T giac CNPQ la hnh thang, co H la trung iem PQ, hai canh ben NP va CQ ong quy tai A nen K la trung iem CN MK la ng trung bnh cuaABCN MK // CN MK // AB (1) H la trc tam cuaAABC nen CHA B (2) T (1) va (2) suy ra MKCH MK la ng cao cuaACHK (3) TAHBC MCHK MI la ng cao cuaACHK(4) T (3) va (4) suy ra M la trc tam cuaACHK MHCN MHPQ3) bai 3: Cho hnh ch nhat ABCD co M, N th t la trung iem cua AD, BC. Goi E la mot iem bat ky thuoc tia oi cua tia DC, K la giao iem cua EM va AC. Chng minh rang: NM la tia phan giac cua

KNEGiaiGoi H la giao iem cua KN va DC, giao iem cua AC va MN la I th IM = IN Ta co: MN // CD (MN la ng trung bnh cua hnh HGFENMDCBAIKNMQPHC BA// //IH ENMKD CBAwww.VNMATH.comwww.vnmath.com 54ch nhat ABCD) T giac EMNH la hnh thang co hai canh ben EM va HN ong quy tai K va I la trung iem cua MN nen C la trung iem cua EH TrongAENH th NC va la ng cao, va la ng trung tuyen nenAENH can tai N NC la tia phan giac cua

ENH ma NCMN (Do NMBC MN // AB) NM la tia phan giac goc ngoai tai N cuaAENH Vay NM la tia phan giac cua

KNEBai 4: Tren canh BC = 6 cm cua hnh vuong ABCD lay iem E sao cho BE = 2 cm. Tren tia oi cua tia CD lay iem F sao cho CF = 3 cm. Goi M la giao iem cua AE va BF.Tnh

AMC Giai Goi giao iem cua CM va AB la H, cua AM va DF la G Ta co: BH AB BH 6 = CF FG 3 FG = Ta lai co AB BE 2 1 = =CG = 2AB = 12 cmCG EC 4 2= FG = 9 cm BH 6BH = 2 cm3 9= BH = BE ABAE =ABCH (c.g.c) BAE = BCH ma BAE + BEA= 900

Mat khac BEA = MEC; MCE = BCH MEC + MCE = 900

AMC = 900 Bai 5: Cho t giac ABCD. Qua iem E thuoc AB, H thuoc AC ve cac ng thang song song vi BD, cat cac canh con lai cua t giac tai F, G a) Co the ket luan g ve cac ng thang EH, AC, FG b) Goi O la giao iem cua AC va BD, cho biet OB = OD. Chng minh rang ba ng thang EG, FH, AC ong quy Giai a) Neu EH // AC th EH // AC // FG Neu EH va AC khong song song th EH, AC, FG ong quy b) Goi giao iem cua EH, HG vi AC Trong hnh thang DFEB co hai canh ben DF, BE ong quy tai A va OB = OD nen theo bo e hnh thang th M la trung iem cua EF Tng t: N la trung iem cua GH Ta co ME MF = GN HNnen ba ng thang EG, FH, AC ong quy tai O HMG FED CB AOHGFENMD CBAwww.VNMATH.comwww.vnmath.com 55 CHUYEN E 14 S DUNG CONG THCDIEN TCH E THIET LAP QUAN HE O DAI CUA CAC OAN THANG A. Mot so kien thc: 1. Cong thc tnh dien tch tam giac: S = 12 a.h (a o dai mot canh, h o dai ng cao tng ng) 2. Mot so tnh chat: Hai tam giac co chung mot canh, co cung o dai ng cao th co cung dien tch Hai tam giac bang nhau th co cung dien tch B. Mot so bai toan: 1. Bai 1: ChoAABC co AC = 6cm; AB = 4 cm; cac ng cao AH; BK; CI. BietAH = CI + BK2 Tnh BC Giai Ta co:BK = ABC2SAC ; CI = ABC2SAB BK + CI = 2. SABC 1 1AC AB| |+ |\ .

2AH = 2.12. BC. AH . 1 1AC AB| |+ |\ .BC.1 1AC AB| |+ |\ . = 2 BC = 2 : 1 1AC AB| |+ |\ . = 2 :1 16 4| |+ |\ . = 4,8 cm Bai 2: ChoAABC co o dai cac canh la a, b, c; o dai cac ng cao tng ng la ha, hb, hc. Biet rang a + ha = b + hb = c + hc . Chng minh rangAABC la tam giac eu Giai Goi SABC = S Ta xet a + ha = b + hb a b = ha hb = 2S 2S 1 1 a - b -2S.-2S. b a b a ab| |= = |\ . a b = a - b2S. ab (a b) 2S1 - ab| | |\ . = 0AABC can C hoac vuong C (1) Tng t ta co:AABC can A hoac vuong A (2);AABC can B hoac vuong B (3) T (1), (2) va (3) suy raAABC can hoac vuong ba nh (Khong xay ra vuong tai ba nh)AABC la tam giac eu Bai 3: Cho iem O nam trong tam giac ABC, cac tia AO, BO, Co cat cac canh cua tam giac KIH C BAwww.VNMATH.comwww.vnmath.com 56ABC theo th t tai A, B, C. Chng minh rang: a) OA' OB' OC'1AA' BB' CC'+ + =b) OA OB OC2AA' BB' CC'+ + =c) M = OA OB OC6OA' OB' OC'+ + = . Tm v tr cua O etong M co gia tr nho nhat d) N = OA OB OC. . 8OA' OB' OC' =. Tm v tr cua O etch N co gia tr nho nhat Giai Goi SABC = S, S1 = SBOC , S2 = SCOA , S3 = SAOB . Ta co: 3 2 3 2OA'C OA'B 1S S S S OA = = OA' S S S+=(1) OA'C OA'B OA'C OA'B 1AA'C AA'B AA'C AA'BS S S S S OA' = = AA' S S S S S+= =+ (2) T (1) va (2) suy ra2 3S S OA AA' S+=Tng t ta co1 32S S OB OB' S+= ; 1 23S S OC OC' S+=; 2S OB' BB' S=; 3S OC' CC' S=a) 3 1 2S S S OA' OB' OC' S1AA' BB' CC' S S S S+ + = + + = =b) 2 3 1 3 1 2S S S S S S OA OB OC 2S2AA' BB' CC' S S S S+ + ++ + = + + = =c) M = 2 3 1 3 3 3 1 2 1 2 2 11 2 3 2 1 2 3 3 1S S S S S S S S S S S S OA OB OCOA' OB' OC' S S S S S S S S S| | | | | | + + ++ + = + + = + + + + + |||\ . \ . \ . Ap dung Bt Co si ta co3 3 1 2 2 12 1 2 3 3 1S S S S S S2 2 2 6S S S S S S| | | | | |+ + + + + > + + = |||\ . \ . \ .

ang thc xay ra khi S1 = S2 = S3 O la trong tam cua tam giac ABC d) N = ( )( )( )2 3 1 3 1 2 2 3 1 3 1 21 2 3 1 2 3S S S S S S S S S S S S. .S S S S .S .S+ + + + + += N2 =( ) ( ) ( )( ) ( )2 2 22 3 1 3 1 2 1 2 2 3 1 32 21 2 3 1 2 3S S S S S S 4S S .4S S .4S S64S .S .S S .S .S+ + +> > N> 8 ang thc xay ra khi S1 = S2 = S3 O la trong tam cua tam giac ABC Bai 4: Cho tam giac eu ABC, cac ng caoAD, BE, CF; goi A, B, C la hnh chieu cua M(nam ben trong tam giac ABC) tren AD, BE, CF. Chng minh rang: Khi M thay oi v tr trong tam giac ABC th: a) AD + BE + CF khong oi b) AA + BB + CC khong oi Giai C'B'A'OCBAwww.VNMATH.comwww.vnmath.com 57Goi h = AH la chieu cao cua tam giac ABC th h khong oi Goi khoang cach t M en cac canh AB; BC; CA la MP; MQ; MRth AD + BE + CF = MQ + MR + MP V M nam trong tam giac ABC nen SBMC + SCMA + SBMA = SABC

BC.(MQ + MR + MP) = BC.AH MQ + MR + MP = AH AD + BE + CF = AH = h Vay: AD + BE + CF = AH = h khong oi b) AA + BB + CC = (AH AD)+(BE BE) (CF CF) = (AH + BE + CF) (AD + BE + CF) = 3h h = 2h khong oi Bai 5:Cho tam giac ABC co BC bang trung bnh cong cua AC va AB; Goi I la giao iem cua cac phan giac, G la trong tam cua tam giac. Chng minh: IG // BC Giai Goi khoang cach t a, I, G en BC lan lt la AH, IK, GD V I la giap iem cua ba ng phan giac nen khoang cach t I en ba canh AB, BC, CA bang nhau va bang IK V I nam trong tam giac ABC nen: SABC = SAIB + SBIC + SCIABC.AH = IK(AB+BC+CA) (1) MaBC = AB + CA 2 AB + CA = 2 BC (2) Thay (2) vao (1) ta co: BC. AH = IK. 3BC IK = 13AH (a) V G la trong tam cua tam giac ABC nen: SBGC = 13 SABC BC . GD = 13 BC. AH GD = 13 AH (b) T (a) va (b) suy ra IK = GD hay khoang cach tI, G en BC bang nhau nen IG // BCBai tap ve nha: 1) Cho C la iem thuoc tia phan giac cua

0xOy = 60 , M la iem bat ky nam tren ng vuong goc vi OC tai C va thuoc mien trong cua

xOy , goi MA, MB th t la khoang cach t M en Ox, Oy. Tnh o dai OC theo MA, MB 2) Cho M la iem nam trong tam giac eu ABC. A, B, C la hnh chieu cua M tren cac canh BC, AC, AB. Cac ng thang vuong goc vi BC tai C, vuong goc vi CA tai A , vuong goc vi AB tai B cat nhau D, E, F. Chng minh rang: a) Tam giac DEF la tam giac eu b) AB + BC + CA khong phu thuoc v tr cua M trong tam giac ABC RQPC'B'A'MFED C BAM K HG IDC BAwww.VNMATH.comwww.vnmath.com 58CHUYEN E 15 TM GIA TR LN NHAT, NHO NHAT CUA MOT BIEU THC A. Gia tr ln nhat, gia tr nho nhat cua mot bieu thc: 1) Khai niem: Neu vi moi gia tr cua bien thuoc mot khoang xac nh nao o ma gia tr cua bieu thc A luon luon ln hn hoac bang (nho hn hoac bang) mot hang so k va ton tai mot gia tr cua bien e A co gia tr bang k th k goi la gia tr nho nhat (gia tr ln nhat) cua bieu thc A ng vi cac gia tr cua bien thuoc khoang xac nh noi tren 2) Phng phapa) e tm gia tr nho nhat cua A, ta can: + Chng minh A> k vi k la hang so + Ch ra da = co the xay ra vi gia tr nao o cua bien b) e tm gia tr ln nhat cua A, ta can: + Chng minh As k vi k la hang so + Ch ra da = co the xay ra vi gia tr nao o cua bien K hieu : min A la gia tr nho nhat cua A; max A la gia tr ln nhat cua A B.Cac bai tap tm Gia tr ln nhat, gia tr nho nhat cua mot bieu thc: I) Dang 1: Tam thc bac hai V du 1 : a) Tm gia tr nho nhat cua A = 2x2 8x + 1 b) Tm gia tr ln nhat cua B = -5x2 4x + 1 Giai a) A = 2(x2 4x + 4) 7 = 2(x 2)2 7 > - 7min A= - 7 x = 2 b) B = - 5(x2 + 45x) + 1 = - 5(x2 + 2.x.25 + 425) + 95 = 95 - 5(x + 25)2s 95 max B = 95 x = 25b) V du 2:Cho tam thc bac hai P(x) = a x2 + bx + c a) Tm min P neua > 0b) Tm max P neu a < 0 Giai Ta co:P = a(x2 + b ax) + c = a(x + b 2a)2 + (c - 2b4a) at c - 2b4a = k. Do (x + b 2a)2> 0 nen: a) Neu a > 0 tha(x + b 2a)2> 0 do o P> k min P = k x =- b 2a b) Neu a < 0 tha(x + b 2a)2s 0 do o Ps k max P = k x =- b 2a www.VNMATH.comwww.vnmath.com 59II. Dang 2: a thc co dau gia tr tuyet oi 1) V du 1: Tm gia tr nho nhat cua a)A = (3x 1)2 4 3x - 1 + 5 at 3x - 1 = y th A = y2 4y + 5 = (y 2)2 + 1> 1 min A = 1 y = 2 3x - 1 = 2 x = 13x - 1 = 213x - 1 = - 2 x = - 3

b) B = x - 2 + x - 3 B = x - 2 + x - 3 = B = x - 2 +3 - x>x - 2+ 3 - x = 1 min B= 1 (x 2)(3 x)> 0 2s xs 3 2) V du 2: Tm GTNN cua C = 2 2 x- x + 1 x- x - 2+Ta co C = 2 2 x- x + 1 x- x - 2+= 2 2 2 2x- x + 1 2 + x - x x- x + 1 + 2 + x - x + >= 3 min C = 3(x2 x + 1)(2 + x x2)> 0 2 + x x2 > 0 x2 x 2s 0 (x + 1)(x 2)s 0- 1 x 2 s s3) V du 3:Tm gi tr nh nht ca :T = |x-1| + |x-2| +|x-3| + |x-4| Ta c |x-1| + |x-4|=|x-1| + |4-x| >|x-1+4-x|=3 (1) V 2 3 2 3 2 3 x x x x x x + = + > + = 1 (2) Vy T = |x-1| + |x-2| +|x-3| + |x-4| >1 + 3 = 4 Ta c t(1)Du bng xy ra khi 1 4 x s s (2)Du bng xy ra khi2 3 x s s Vy T c gi tr nh nht l 4 khi 2 3 x s sIII.Dang 3: a thc bac cao 1) V du 1: Tm gia tr nho nhat cua a) A = x(x 3)(x 4)(x 7) = (x2 7x)( x2 7x + 12) atx2 7x + 6 th A = (y 6)(y + 6) = y2 36> - 36 Min A = - 36 y = 0 x2 7x + 6 = 0(x 1)(x 6) = 0x = 1 hoac x = 6 b) B = 2x2 + y2 2xy 2x + 3 = (x2 2xy + y2) + (x2 2x + 1) + 2 = (x y)2 + (x 1)2 + 2> 2 x - y = 0 x = y = 1x - 1 = 0 c) C = x2 + xy + y2 3x 3y = x2 2x + y2 2y + xy x y Ta coC + 3 = (x2 2x + 1) + (y2 2y + 1) + (xy x y + 1)= (x 1)2 + (y 1)2 + (x 1)(y 1). at x 1 = a; y 1 = b th C + 3 = a2 + b2 + ab = (a2 + 2.a.b2 + 2b4) + 23b4 = (a + b2)2 + 23b4> 0 Min (C + 3) = 0 hay min C = - 3 a = b = 0 x = y = 1 2) V du 2:Tm gia tr nho nhat cua a) C = (x + 8)4 + (x + 6)4

www.VNMATH.comwww.vnmath.com 60at x + 7 = y C = (y + 1)4 + (y 1)4 = y4 + 4y3 + 6y2 + 4y + 1 + y4 - 4y3 + 6y2 - 4y + 1 = 2y4 + 12y2 + 2> 2 min A = 2 y = 0 x = - 7 b) D = x4 6x3 + 10x2 6x + 9 = (x4 6x3 + 9x2 ) + (x2 6x + 9)= (x2 3x)2 + (x 3)2> 0 min D = 0 x = 3 IV. Dang phan thc: 1. Phan thc co t la hang so, mau la tam thc bac hai Bieu thc dang nay at GTNN khi mau at GTLN V du : Tm GTNN cuaA = 226x - 5 - 9x = 2 2- 2 29x - 6x + 5(3x - 1) 4=+ V(3x 1)2> 0 (3x 1)2 + 4> 4 2 21 1 2 2(3x - 1) 4 4 (3x - 1) 4 4 s >+ + A> - 12 min A = -12 3x 1 = 0 x = 13 2. Phan thc co mau la bnh phng cua mot nh thc a) V du 1: Tm GTNN cuaA = 223x- 8x + 6x- 2x + 1 +) Cach 1: Tach t thanh cac nhom co nhan t chung vi mau A = 2 22 2 23x- 8x + 6 3(x- 2x + 1) - 2(x - 1) + 1 2 1 =3x- 2x + 1 (x - 1) x - 1 (x - 1)= + . at y = 1x - 1 Th A = 3 2y + y2 = (y 1)2 + 2> 2 min A = 2 y = 1 1x - 1 = 1 x = 2 +) Cach 2: Viet bieu thc A thanh tong cua mot so vi mot phan thc khong am A = 2 2 2 22 2 23x- 8x + 6 2(x- 2x + 1)+ (x- 4x + 4) (x - 2) =2 2x- 2x + 1 (x - 1) (x - 1)= + > min A = 2 x 2 = 0 x = 2 b) V du 2: Tm GTLN cua B = 2xx 20x + 100 + Ta co B = 2 2x xx 20x + 100 (x + 10)=+. at y = 1x + 10 x = 110y th B = (110y ).y2 = - 10y2 + y = - 10(y2 2.y.120y + 1400) + 140 = - 1021y - 10| | |\ .+ 140 s 140 Max B = 140 1y - 10 = 0 y = 110 x = 10 c) V du 3: Tm GTNN cuaC = 2 22 2x+ yx+ 2xy + y Ta co: C = 2 22 2 22 2 2 21(x + y) (x - y)x+ y 1 1 (x - y) 12.x+ 2xy + y (x + y) 2 2 (x + y) 2 ( + = = + > min A = 12 x = y 3. Cac phan thc co dang khac a)V du : Tm GTNN, GTLN (Cc tr) cuaA = 23 - 4xx 1 + www.VNMATH.comwww.vnmath.com 61Ta co:A = 2 2 22 2 23 - 4x (4x 4x4) (x 1) (x - 2)1 1x 1 x 1 x 1 + += = > + + + min A = - 1 x = 2 Ta lai co: A = 2 2 22 2 23 - 4x (4x 4) (4x+ 4x + 1)(2x 1)4 4x 1 x 1 x 1+ += = s+ + + max A = 4 x = 12C. Tm GTNN, GTLN cua mot bieu thc biet quan he gia cac bien 1) V du 1:Cho x + y = 1. Tm GTNN cua A = x3 + y3 + xy Ta co A = (x + y)(x2 xy + y2) + xy = x2 + y2 (v x + y = 1) a) Cach 1:Bieu th an nay qua an kia, roi a ve mot tam thc bac hai Tx + y = 1 x = 1 ynen A = (1 y)2 + y2 =2(y2 y) + 1 = 2(y2 2.y.12 + 14) + 12 = 221 1 1y - + 2 2 2| |> |\ . Vay min A = 12 x =y = 12 b) Cach 2: S dung k a cho, lam xuat hien mot bieu thc mi co cha A T x + y = 1 x2 + 2xy + y2 = 1(1). Mat khac (x y)2> 0 x2 2xy + y2> 0 (2) Cong (1) vi (2) ve theo ve, ta co: 2(x2 + y2)> 1 x2 + y2> 12 min A = 12 x = y = 12 2)V du 2:Cho x + y + z = 3 a) Tm GTNN cua A = x2 + y2 + z2 b) Tm GTLN cua B = xy + yz + xz TCho x + y + z = 3 Cho (x + y + z)2 = 9 x2 + y2 + z2 + 2(xy + yz + xz) = 9 (1) Ta co x2 + y2 + z2- xy yz zx= 21.2 .( x2 + y2 + z2- xy yz zx) =212 2 2( ) ( ) ( ) x y x z y z( + + > 0 x2 + y2 + z2>xy+ yz + zx (2) ang thc xay ra khix = y = za) T (1) va (2) suy ra9 = x2 + y2 + z2 + 2(xy + yz + xz)s x2 + y2 + z2 + 2(x2 + y2 + z2) = 3(x2 + y2 + z2) x2 + y2 + z2> 3 min A = 3 x = y = z = 1 b) T (1) va (2) suy ra 9 = x2 + y2 + z2 + 2(xy + yz + xz)> xy+ yz + zx + 2(xy + yz + xz) = 3(xy+ yz + zx) xy+ yz + zxs 3 max B = 3 x = y = z = 1 3) V du 3:Tm gi tr ln nht caS = xyz.(x+y).(y+z).(z+x) vi x,y,z > 0v x + y + z= 1 V x,y,z > 0 ,p dng BT Csi ta c: x+ y + z 33 xyz >31 13 27xyz xyz s sp dng bt ng thc Csi cho x+y ; y+z;x+z ta c( ) ( ) ( ) ( ) ( ) ( )3. . 3 . . x y y z z x x y y z x z + + + > + + +( ) ( ) ( )32 3 . . x y y z z x > + + + Du bng xy ra khix = y = z =13Ss 8 1 8.27 27 729=www.VNMATH.comwww.vnmath.com 62VyS c gi tr ln nht l8729 khi x = y = z = 13 4) V du 4:Cho xy + yz + zx = 1. Tm gi tr nh nht ca4 4 4x y z + +p dng BT Bunhiacpski cho 6 s(x,y,z) ;(x,y,z) Ta c( ) ( )222 2 2xy yz zx x y z + + s + +( )22 2 21 x y z s + + (1) p dng BT Bunhiacpski cho (2 2 2, , x y z ) v (1,1,1) Ta c2 2 2 2 2 2 2 4 4 4 2 2 2 2 4 4 4( ) (1 1 1 )( ) ( ) 3( ) x y z x y z x y z x y z + + s + + + + + + s + + T (1) v (2) 4 4 41 3( ) x y z s + +4 4 413x y z + + sVy4 4 4x y z + +c gi tr nh nht l13 khix= y = z = 33D. Mot so chu y: 1) Khi tm GTNN, GTLN ta co the oi bien V du : Khi tm GTNN cua A =(x 1)2 + (x 3)2 , ta at x 2 = y thA = (y + 1)2 + (y 1)2 = 2y2 + 2> 2 2) Khi tm cc tr cua mot bieu thc, ta co the thay k cua bieu thc nay at cc tr bi k tng ng la bieu thc khac at cc tr:+) -A ln nhat A nho nhat ;+) 1Bln nhat B nho nhat (vi B > 0) +) C ln nhat C2 ln nhat V du: Tm cc tr cua A = ( )422x+ 1x+ 1 a) Ta co A > 0 nen A nho nhat khi 1A ln nhat, ta co ( )2224 4x+ 11 2x1 1A x+ 1 x+ 1= = + > min 1A = 1 x = 0 max A = 1 x = 0 b) Ta co (x2 1)2> 0 x4 - 2x2 + 1>0x4 + 1> 2x2. (Dau bang xay ra khi x2 = 1) V x4 + 1 > 0 242xx+ 1s 1 242x1 1 1 2x+ 1+ s + = max 1A = 2 x2 = 1 min A = 12 x = 1 3) Nhieu khi ta tm cc tr cua bieu thc trong cac khoang cua bien, sau o so samh cac cc tr o e e tm GTNN, GTLN trong toan bo tap xac nh cua bien V du: Tm GTLN cua B = y 5 - (x + y) a) xet x + ys 4 - Neu x = 0 th A = 0 - Neu 1 y 3 s sth As 3 - Neu y = 4 th x = 0 va A = 4 b) xet x + y> 6 th As 0 www.VNMATH.comwww.vnmath.com 63So sanh cac gia tr tren cua A, ta thay max A = 4 x = 0; y = 4 4) S dung cac hang bat ang thc V du: Tm GTLN cua A = 2x + 3y biet x2 + y2 = 52 Ap dung Bt Bunhiacopxki: (a x + by)2s (a2 + b2)(x2 + y2) cho cac so2, x , 3, y ta co: (2x + 3y)2s (22 + 32)(x2 + y2) = (4 + 9).52 = 2622x + 3ys 26 Max A = 26 x y = 2 3 y = 3x2 x2 + y2 = x2 + 23x2| | |\ . = 52 13x2 = 52.4 x =4 Vay: Ma x A = 26 x = 4; y = 6 hoac x = - 4; y = - 6 5) Hai so co tong khong oi th tch cua chung ln nhat khi va ch khi chung bang nhau Hai so co tch khong oi th tong cua chung ln nhat khi va ch khi chung bang nhau a)V du 1:Tm GTLN cua A = (x2 3x + 1)(21 + 3x x2) V(x2 3x + 1) + (21 + 3x x2) = 22 khong oi nen tch (x2 3x + 1)(21 + 3x x2) ln nhat khi va ch khi x2 3x + 1 = 21 + 3x x2 x2 3x 10 = 0 x = 5 hoac x = - 2 Khi o A = 11. 11 = 121 Max A = 121 x = 5 hoac x = - 2 b) V du 2: Tm GTNN cua B = (x + 4)(x + 9)x Ta co: B = 2(x + 4)(x + 9) x 13x + 36 36x +13x x x+= = + V cac so x va 36x co tchx.36x = 36 khong oi nen 36x + x nho nhatx = 36x x = 6 A = 36x +13x +nho nhat la min A = 25 x = 6 6)Trong khi tm cc tr ch can ch rarang ton tai mot gia tr cua bien e xay ra ang thc ch khong can ch ra moi gia tr e xay ra ang thc V du: Tm GTNN cuaA = m n11 5 Ta thay 11m tan cung bang 1, 5n tan cung bang 5 Neu 11m > 5n th A tan cung bang 6, neu 11m < 5n th A tan cung bang 4 khi m = 2; n = 3 th A =121 124 = 4 min A = 4, chang han khi m = 2, n = 3 www.VNMATH.comwww.vnmath.com 64 GIAI MOT SO E THI Bi 1:a) Thc hin php chia: (x3 - 2x - 4) : (x2 + 2x + 2) b) Xc nh a sao cho ax3 - 2x - 4 chia ht cho x - 2 c) Tm nghim ca a thc: x3 - 2x - 4 Bi 2: a) Tnh S = a b c(c a)(a b) (a b)(b c) (b c)(c a)+ + b) Chng minh 1 1 1 1(3n 2)(3n 5) 3 3n 2 3n 5| |= |+ + + +\ . c) Tnh 150 150 150 150...5.8 8.11 11.14 47.50+ + + +Bi 3: Gii cc phng trnh a) 2 2 4 2x 1 x 1 2x x 1 x x 1 x(x x 1)+ =+ + + + + b) 7 x 5 x 3 x31993 1995 1997 + + = Bi 4:ChoABC Avung ti A. V ra pha ngoi tam gic cc tam gic ABD vung cn B, ACE vung cn C. CD ct AB ti M, BE ct AC ti N a) Chng minh ba im D, A, Ethng hng; cc t gic BCE; ACBD l hnh thang b) Tnh DM bit AM = 3cm; AC = 4 cm; MC = 5cm c) Chng minh AM = AN Bi 5:Cho M l im nm trong ABC A , t M k MA BC, MBAC, MC AB (Ae BC; Be AC; Ce AB). Chng minh rng: a b cMA' MB' MC'h h h+ += 1 (Vi ha, hb, hc l ba ng cao ca tam gic h ln lt t A, B, C xung ba cnh caABC A )Bi gii Bi 1: a) Thc hin php chia: (x3 - 2x - 4) : (x2 + 2x + 2)=x - 2 b) Xc nh a sao cho ax3 - 2x - 4 chia ht cho x - 2Vax3 - 2x - 4 chia ht cho x - 2nn x = 2 l nghim ca a thc ax3 - 2x - 4 , nn ta c: a. 23 - 2. 2 - 4 = 0 8a - 8 = 0 a = 1 c) Tm nghim ca a thc: x3 - 2x - 4 Nghim ca a thc l cc gi tr ca x x3 - 2x - 4 = 0 (x2 + 2x + 2)(x - 2) = 0 2x 2x 2 0x 2 0 + + =

= +)x - 2 = 0 x = 2+)x2 + 2x + 2 (x2 + 2x + 1) + 1 = 0 (x+ 1)2 + 1 = 0 : V nghim V (x+ 1)2 + 1 > 0 vi mi x Bi 2: www.VNMATH.comwww.vnmath.com 65a)S = a b c a(b c) b(c a) c(a b)(c a)(a b) (a b)(b c) (b c)(c a) (c a)(a b)(b c) + + + + = =a(b c) b(c a) c(a b) ab ac bc ab ac bc 00(c a)(a b)(b c) (c a)(a b)(b c) (c a)(a b)(b c) + + + + = = = b) Chng minh 1 1 1 1(3n 2)(3n 5) 3 3n 2 3n 5| |= |+ + + +\ . Ta c:1 1 1 1 3n 5 (3n 2) 1 3 1.3 3n 2 3n 5 3 (3n 2)(3n 5) 3 (3n 2)(3n 5) (3n 2)(3n 5) ( + + | | = = = |(+ + + + + + + +\ . c) Tnh : 150 150 150 150...5.8 8.11 11.14 47.50+ + + + p dngcu b ta tnh c150 150 150 150...5.8 8.11 11.14 47.50+ + + += 9 Bi 3: Gii cc phng trnh a)2 22 2 4 2 4 2 4 2 4 2x 1 x 1 2 x(x 1)(x x 1) x(x 1)(x x 1) 2x x 1 x x 1 x(x x 1) x(x x 1) x(x x 1) x(x x 1)+ + + + + = =+ + + + + + + + + + + (1) KX: x(x4 + x2 + 1)=0 x=0V x4 + x2 + 1 > 0 (1) x(x + 1)(x2 - x + 1) - x(x - 1)(x2 + x + 1) = 2x(x3 - 1) - x(x3 + 1)= 2 x4 - x - x4 - x = 2 - 2x = 2 x = - 1 b)7 x 5 x 3 x 7 x 5 x 3 x3 1 1 1 01993 1995 1997 1993 1995 1997 + + = + + + + + = x = 2000 Bi 4:ChoABC Avung ti A. V ra pha ngoi tam gic cc tam gic ABD vung cn B, ACE vung cn C. CD ct AB ti M, BE ct AC ti N a) Chng minh ba im D, A, Ethng hng; cc t gic BCE; ACBD l hnh thang b) Tnh DM bit AM = 3cm; AC = 4 cm; MC = 5cm c) Chng minh AM = AN Gii a) Chng minh

DAB + BAC + CAE= 1800 D, A, E thng hang b) at AB = c, AC = b.BD // AC (cung vuong goc vi AB)nenMC AM AC AM ACMD MB BD MB + AM AC + BD= = = AM AC AM AC AC. ABAMABAC + BD ABAC AB AC AB= = =+ + (1) AM(AC + AB) = AC. AB 3(4 + AB) = 4 AB AB = 12 cm MB = 9 cm TMC AM MC.MB 5.9MD 15MD MB MA 3= = = =cm c)AB // CE (cung vuong goc vi AC) nen AN AB AN ABNC CE NC + AN AB + CE= = NMEDCBAwww.VNMATH.comwww.vnmath.com 66 AN AB AB. ACANAC AB + AC AB + AC= =(2) T (1) va (2) suy ra: AM= AN Bi 5:Cho M l im nm trong ABC A , t M k MA BC, MBAC, MC AB (Ae BC; Be AC; Ce AB). Chng minh rng: a b cMA' MB' MC'h h h+ += 1 (Vi ha, hb, hc l ba ng cao ca tam gic h ln lt t A, B, C xung ba cnh caABC A )Gii K ng cao AH, ta c: MBCa ABCS MA' MA'h AH S= =(1) Tng t: MCAb ABCS MB'h S=(2) v MBAc ABCS MC'h S=(3) Cng (1), (2) v (3) v theo v, ta c:

MBC MCA MBAa b c ABC ABC ABCS S S MA' MB' MC'h h h S S S+ + = + +=MBC MCA MBA ABCABC ABCS S S S1S S+ += = Cu 1 a) Trong ba s a, b, c c 1 s dng, 1 s m v 1 s bng 0; ngoi ra cn bit thm2a b (b c) = . Hi s no dng, s no m, s no bng 0 b) Cho x + y = 1. Tnh gi tr biu thc A = x3 + y3 + 3xy Cu 2 a) Gii phng trnh:x 2 3 1 + =b) Gi s a, b, c l ba s i mt khc nhau v a b c0b c c a a b+ + = Chng minh rng:2 2 2a b c0(b c) (c a) (a b)+ + = Cu 3: Cho tam gic ABC; gi Ax l tia phn gic ca

BAC, Ax ct BC ti E. Trn tia Ex ly im H sao cho BAE ECH = . Chngminhrng: a) BE. EC = AE. EH b) AE2 = AB. AC - BE. EC Cu 4: Cho t gicABCD. T A k ng thng song song vi BC ct BD ti E; t B k ng thng song song vi AD ct AC ti F.Chng minh rng: EF // DCHC'B'A'MCBAwww.VNMATH.comwww.vnmath.com 67hng dn gii Cu 1: a) V 2a b (b c) = nn a= 0 v b= 0 v Nu a = 0 b = 0 hoc b = c. V lNu b = 0 a = 0. V l c = 0a= b3 ma >0 vi mi a b > 0 a < 0 b) V x + y = 1 A = x3 + y3 + 3xy = x3 + y3 + 3xy (x + y) = (x + y)3 = 1 Cu 2:b) Ta b c +0b - c c - a a - b+ = 2 2a b c b ab + ac - c = b - c a - c b - a (a - b)(c - a)+ = 2 22a b ab + ac - c(b - c) (a - b)(c - a)(b - c)= (1)(Nhan hai ve vi1b - c) Tng t, ta co: 2 22b c bc + ba - a(c - a) (a - b)(c - a)(b - c)=(2) ; 2 22c a ac + cb - b(a - b) (a - b)(c - a)(b - c)=(3) Cong tng ve (1), (2) va (3) ta co pcm Cu 3: a) Ta cABAEAHCE (g.g) BE AEBE.EC AE.EHEH EC= =(1) b) ABAEAHCE (g.g) ABE = CHE ABE = CHA ABAEAHAC (g.g) AE ABAB.AC AE.AHAC AH= =(2) Tr (1) cho (2) v theo v ta c : AB. AC - BE. EC = AE.AH - AE. EH AB. AC - BE. EC = AE. (AH - EH) = AE. AE = AE2 Cu 4:Goi O la giao iem cua AC va BD a) V AE // BC OE OA = OB OC (1) BF // AD OB OF = OD OA (2) Nhan (1) vi (2) ve theo ve ta co: OE OF = OD OC EG // CD Bi 1:Cho phn thc: P = 22 x 4x x 20+ HExCBAOFDECBAwww.VNMATH.comwww.vnmath.com 68a) Tm TX ca P b) Rt gn P c) Tnh gi tr ca P khix 5 1, 5 =Bi 2: So snh A v B bit: a) A = 2002. 2004 v B = 20032 b) A = 3.(22 + 1)(24 + 1)(28 + 1)(216 + 1)(232 + 1) v B = 264 Bi 3:Cho hnh bnh hnh ABCD c ng cho ln AC. H CE vung gc vi AB, CF vung gc vi AD v BG vung gc vi AC. Chng minh: a)AACEAABGvAAFCACBG b) AB. AE + AD. AF = AC2 Bi 4: Cho hnh thoi ABCD cnh a, c = 600. Mt ng thng bt k qua C ct tia i ca tia BA v DA ln lt ti M v N a) Chng minh: Tch BM. DN c gi tr khng i b) Gi K l giao im ca BN v DM. Tnh s o gc BKD Bi 5: Tm nghim nguyn ca phng trnh 4(x + y) = 11 + xy Gii Bi 1: a) kx:x2 + x - 20=0(x - 4)(x + 5)=0 x=4 v x=- 5 b) P = 22 x 4 2 x 4x x 20 (x 4)(x 5) =+ +

Nu x>4 P = 2x 5 + Nu x < 4 P = 2x 5+ c) x 5 1, 5; (x 5) x 6, 5x 5 1, 55 x 1, 5; (x 5) x 3, 5 = > = = = < = Vi x = 6,5 th P = 2 2 2 20 4x 5 6, 5 5 11, 5 115 23= = = =+ + Vi x = 3,5 th P = 2 2 2 2x 5 3, 5 5 8, 5 17 = = =+ + Bi 2: a) A = 2002. 2004 = (2003 - 1)(2003 + 1) = 20032 - 1 < 20032 A < B b) Ta c: A = 3.(22 + 1)(24 + 1)(28 + 1)(216 + 1)(232 + 1) =(22 - 1)(22 + 1)(24 + 1)(28 + 1)(216 + 1)(232 + 1)= (24 - 1)( 24 + 1)(28 + 1)(216 + 1)(232 + 1)= (28 - 1)(28 + 1)(216 + 1)(232 + 1) = (216 - 1)(216 + 1)(232 + 1) = (232 - 1)(232 + 1) = 264 - 1 < 264 A < B Bi 3: www.VNMATH.comwww.vnmath.com 69Ta co AAGBAAECAE AC = AG AB AB. AE = AC. AG (1) ACGBAAFCAF CG CG = AC CB AD= (v CB = AD)AF . AD = AC. CG (2) Cong (5) va (6) ve theo ve ta co:AB. AE + AF. AD = AC. AG + AC. CG AB. AE + AF. AD = AC(AG + CG) = AC. AC Vay:AB. AE + AD. AF = AC2 Bi4: a) BC // AN MB CM = BA CN(1) CD// AM CM AD = CN DN (2) T (1) va (2) suy ra 2MB AD =MB.DN = BA.AD = a.a = aBA DN b)AMBD va ABDN co

MBD = BDN= 1200 MB MB CM AD BD = = BD BA CN DN DN= = (Do ABCD la hnh thoi co

0A = 60 nenAB = BC = CD = DA) AMBDABDN Suy ra 1 1 M= B .AMBD va ABKD co BDM = BDK va 1 1 M= Bnen 0BKD = MBD= 120 Cu 1:Cho 22x 7x 6Ax 1 += a) Rt gn A b) Tm x A = 0 c) Tm gi tr nguyn ca x A c gi tr nguyn Cu 2: Gii phng trnh: (x + 1)2 = 4(x2 + 2x + 1) Cu 3: Cho a, b, c tho mn: 1 1 1 1a b c a b c+ + =+ + Tnh gi tr ca biu thc: A = (a3 + b3)(b3 + c3)(c3 + a3) Cu 4:Cho AABC c

A 2B 4C 4 = = = o . Chng minh: 1 1 1AB BC CA= +Cu 5: ChoAABC cn ti A c BC = 2a, M l trung im ca BC. Ly D, E theo th t thuc AB, AC sao cho: DME B =a) Chng minh rng: tch BD. CE khng i b) Chng minh rng DM l tia phn gic ca gc BDE c) Tnh chu vi caAADEnuAABCl tam gic u www.VNMATH.comwww.vnmath.com 70Hng dn Cu 3:T1 1 1 1a b c a b c+ + =+ + 1 1 1 10a b c a b c+ + - =+ + a b a b0ab c(a b c)+ ++ =+ + c(a b c) ab(a b). 0 (a + b)(b + c)(c + a) = 0abc(a b c)+ + ++ = + +

T suy ra : A = (a3 + b3)(b3 + c3)(c3 + a3) = ( a + b)(b + c)(c + a). B = 0 Cu 4 :V tia CM (M e AB) sao cho

ACM = o CAM AvCBM Al cc tam gic cn AB AB AM AB AM AB BM1BC AC CM CM CM CM++ = + = = =(v BM = CM) AB AB 1 1 11BC AC AB BC CA+ = = +Cu 5 : a) Ta co

DMC = DME + CME = B + BDM, ma DME = B(gt) nen CME = BDM, ket hp vi B = C ( AABCcan tai A) suy raABDMACME (g.g) 2BD BM = BD. CE = BM. CM = aCM CE khong oi b)ABDMACME DM BD DM BD = = ME CM ME BM(do BM = CM) ADMEADBM (c.g.c) MDE = BMD hay DM la tia phan giac cua

BDEc) chng minh tng t ta co EM la tia phan giac cua

DEC keMHCE ,MIDE, MKDB th MH = MI = MKADKM =ADIMDK =DIAEIM =AEHMEI = EH Chu viAAED la PAED = AD + DE + EA = AK +AH = 2AH (V AH = AK) A ABC la tam giac eu nen suy ra CH = MC2 2a= AH = 1,5a PAED = 2 AH = 2. 1,5 a = 3a

5 - kho st cht lng hc sinh giilc h(2009 - 2010) Cu 1 : Gii phng trnh :a) ) 4 ( . ) 2 (24321x x xxxx +++ b)6x2 - x - 2 = 0 Cu 2 :Cho x + y + z = 0. Rt gn : 2 2 22 2 2) ( ) ( ) ( y x x z z yz y x + + + + Cu 3 : Chng minh rng khng tn ti x tha mn : a)2x4 - 10x2 + 17 = 0 b)x4 - x3 + 2x2 - x + 1 = 0 3o4oo3o2ooMCBAKHIMEDC BAwww.VNMATH.comwww.vnmath.com 71Cu 4 : Cho tam gic ABC, im D nm trn cnh BC sao cho 21=DCDB ; im O nm trn on AD sao cho OA 3OD 2= . Gi K l giao im ca BO v AC.Tnh t s AK : KC. Cu 5 : Cho tam gic ABC c 3 gc nhn, trc tm H. Mt ng thng qua H ct AB, AC th t P v Q sao cho HP = HQ. Gi M l trung im ca BC. Chng minh rng tam gic MPQ cn ti M. hng dn gii Cu 2: Tx + y + z = 0 x2 + y2 + z2 = - 2(xy + yz + zx) (1) Ta c: (x - y)2 + (y - z)2 + (z - x)2 = 2(x2 + y2 + z2 ) - 2(xy + yz + zx)(2) T (1) v (2) suy ra: (x - y)2 + (y - z)2 + (z - x)2 = - 6(xy + yz + zx)(3) Thay (1) v (3) vo biu thc A ta c: A = - 2(xy + yz + zx)1- 6(xy + yz + zx)3=Cu 3: a) 2x4 - 10x2 + 17 = 0 2( x4 - 5x2 + 172) = 02(x4 - 2. 52 x2 + 254)2 +92 = 0 2(x2 - 52)2 + 92 = 0. V 2(x2 - 52)2 + 92 > 0 vi mi x nn khng tn ti x 2x4 - 10x2 + 17 = 0 b) x4 - x3 + 2x2 - x + 1 = 0(x2 + 1)(x2 - x + 1) = 0 V v phi lun dng vi mi x nn khng tn ti x x4 - x3 + 2x2 - x + 1 = 0Cu 4: T D k DM // BK. p dng nh l Talt voAAOK ta c: AK AO 3KM OD 2= =(1) Tng t, trongACKB th: KM CD 1CK DB 3= =(2) Nhn (1) vi (2) v theo v ta c: AK 1CK 2=Cu 5 Goi giao iem cua AH va BC la I T C ke CN // PQ (Ne AB),T giac CNPQ la hnh thang, co H la trung iem PQ, hai canh ben NP va CQ ong quy tai A nen K la trung iem CN MK la ng trung bnh cuaABCN MK // CN MK // AB (1) H la trc tam cuaAABC nen CHA B (2) T (1) va (2) suy ra MKCH MK la ng cao cuaACHK (3) OKMCD BAIKNMQPHC BAwww.VNMATH.comwww.vnmath.com 72TAHBC MCHK MI la ng cao cuaACHK(4) T (3) va (4) suy ra M la trc tam cuaACHK MHCN MHPQAMPQ c MH va l ng trung tuyn va l ng cao nn cn ti M 6 -thi HSG Ton 8 - cp huyn Cu 1:a) Tm cc s nguyn m, n tho mn2n n 1mn 1+ +=+

b) t A = n3 + 3n2 + 5n + 3 . Chng minh rng A chia ht cho 3 vi mi gitr nguyn dng ca n. c) Nu a chia 13 d 2 v b chia 13 d 3 th a2+b2 chia ht cho 13. Cu2 : Rt gn biu thc:a)A= ) )( ( c a b abc + ) )( ( a b c bca + ) )( ( b c a cab b)B = 6 36 36 31 1 1 1x x 2 : x xx x x x ((| | | | | |+ + + + + (( |||\ . \ . \ . (( Cu 3:Tnh tng:S = 3 . 11 + 5 . 31 + 7 . 51 + + 12009.2011 Cu 4: Cho 3 s x, y, z,tho mn iu kin xyz = 2011. Chng minh rng biu thc sau khng ph thuc vo cc bin x, y, z : 2011x y zxy 2011x 2011 yz y 2011 xz z 1+ ++ + + + + + Cu 5: Gii phng trnh:69 x 67 x 65 x 63 x 61 x51942 1944 1946 1948 1950 + + + + = Cu 6: ChoAABC tam gic u, gi M l trung im ca BC . Mt gc

xMy= 600 quay quanh im M sao cho 2 cnh Mx , My lu