Chng 1: Mnh -Tp hp1 1. Mnh v mnh cha bin1. Mnh mnh cha bina) Mnh Mnh lgic (gi tt l mnh ) l mt cu khng nh hoc ng hoc sai. Mt mnh khng th va ng va sai.Mt cu khng nh ng gi l mnh ng. Mt cu khng nh sai gi l mnh sai. V d 1: a) Gc vung c s o 800 (l mnh sai)b) S 7 l mt s nguyn t (l mnhng)c) Hm nay tri p qu ! (khng l mnh )d) Bn c khe khng ? (khng l mnh )V d 2: Trong cc cu sau y cu no l mnh ? Nu l mnh hy xc nh xem mnh ng hay sai.a) Khng c i li ny!b) By gi l my gi?c) Chin tranh th gii ln th hai kt thc nm 1946.d) 16 chia 3 d 1.f) 2003 khng l s nguyn t.e)5 l s v t.Ch :+ Cc cu hi, cu cm thn, cu mnh lnh khng phi l mnh .+ Mnh thng c k hiu bng cc ch ci in hoa.V d:Q: 36 chia ht cho 12+ Mt cu m cha th ni ng hay sai nhng chc chn n ch ng hoc sai, khng th va ng va sai cng l mnh .V d: C s sng ngoi Tri t l mnh .b) Mnh cha binNhng cu khng nh m tnh ng-sai ca chng ty thuc vo gi tr ca bin c gi l nhng mnh cha bin.V d: Cho P(x): x > x2 vi x l s thc. Khi :P(2) l mnh sai, P(1/2) l mnh ng.2. Mnh ph nhCho mnh P. Mnh Khng phi P c gi l mnh ph nh ca P v khiu lP. Mnh P ng nu P sai vP sai nu P ng.Ch : Mnh ph nh ca P c th din t theo nhiu cch khc nhauV d: P: 5 l s v t. Khi mnh Pc th pht biu : 5 khng phi l s v t hoc 5 l s hu t.3. Mnh ko theo +Cho hai mnh P v Q. Mnh Nu P th Q c l mnh ko theo+K hiu l PQ. + Mnh ko theo ch sai khi P ng Q sai.* PQ cn c pht biu l P ko theo Q,P suy ra Q hay V P nn QV d: Cho t gic ABCD.Xt hai mnh P : T gic ABCD l mt hnh ch nht Q : T gic ABCD l mt hnh bnh hnh PQ: Nu t gic ABCD l hnh ch nht th t gic ABCD lhnh bnh hnh .-1-QP Nu t gic ABCD l hnh bnh hnh th t gic ABCD l hnh ch nht .* Trong ton hc, nh l l mt mnh ng, thng cdng : PQ P gi l gi thit, Q gi l kt lun. HocP(x) l iu kin c Q(x)Q(x) l iu kin cn c P(x)Hoc iu kin c Q(x) l P(x)iu kin cn c P(x) l Q(x)4. Mnh o-Mnh tng nga) Mnh o: Cho mnh PQ. Mnh QP c gi l mnh o ca PQb) Mnh tng ng+ Mnh P nu v ch nu Q (P khi v ch khi Q) c gi l mnh tng ng,+ K hiu PQ +Mnh PQ ng khi PQ ng v QP ng v sai trong cc trng hp cn li.( hay PQ ng nu c hai P v Q cng ng hoc cng sai)Cc cch c khc:P tng ng QP l iu kin cn v c Qiu kin cn v c P(x) l c Q(x) V d 1: Xt cc mnh A: 36 chia ht cho 4 v chia ht cho 3;B: 36 chia ht 12Khi : A ng; B ngAB: 36 chia ht cho 4 v chia ht cho 3 nu v ch nu 36 chia ht 12. ngV d 2: Mnh Tam gic ABC l tam gic c ba gc bng nhau nu v ch nu tam gic c ba cnh bng nhau l mnh g? Mnh ng hay sai? Gii thch.XtP: Tam gic ABC l tam gic c ba gc bng nhauQ: Tam gic c ba cnh bng nhauKhi PQ ng; QP ng. Vy PQ6. Cc k hiu v K hiu (vi mi): ) ( , " x P X x hoc ) ( : x P X x K hiu (tn ti) : ) ( , x P X x hoc ) ( : x P X x Ph nh ca mnh x X, P(x) l mnh xX,P(x)Ph nh ca mnh x X, P(x) l mnh xX,P(x)V d: Cc bit tnh ng/sai ca cc mnh sau? Nu mnh ph nh.a) n *, n2-1 l bi ca 3b) x , x2-x+1>0c) x , x2=3d) n , 2n + 1 l s nguyn te) n , 2n n+2.* Trong ton hc, nh l l mt mnh ng, thng cdng : PQ P gi l gi thit, Q gi l kt lun. HocP(x) l iu kin c Q(x)Q(x) l iu kin cn c P(x)Hoc iu kin c Q(x) l P(x)iu kin cn c P(x) l Q(x)-2-* Mnh tng ng+ Mnh P nu v ch nu Q (P khi v ch khi Q) c gi l mnh tng ng.K hiu PQ +Mnh PQ ng khi PQ ng v QP ng v sai trong cc trng hp cn li.( hay PQ ng nu c hai P v Q cng ng hoc cng sai)Cc cch c khc:P tng ng QP l iu kin cn v c Qiu kin cn v c P(x) l c Q(x).B sung:Trong lgic ton, mt phn ngnh lgic hc, c s ca mi ngnh ton hc,mnh , hay gi y l mnh lgic l mt khi nim nguyn thy, khng nh ngha.Ch :(mnh )1. Trong thc t c nhng mnh m tnh ng sai ca n lun gn vi mt thi gian v a im c th: ng thi gian hoc a im ny nhng sai thi gian hoc a im khc. Nhng bt k thi im no, a im no cng lun c gi tr chn l ng hoc sai. V d: Sng nay bn An i hc.Tri ma.Hc sinh tiu hc ang i ngh h.2. Ta tha nhn cc lut sau y ca lgic mnh : Lut bi trng: Mi mnh phi hoc ng, hoc sai; khng c mnh no khng ng cng khng sai.Lut mu thun: Khng c mnh no va ng li va sai.3. C nhng mnh m ta khng bit (hoc cha bit) ng hoc sai nhng bit "chc chc" n nhn mt gi tr. V d: Trn sao Ha c s sng.Ch :(mnh ko theo)1. Trong lgic, khi xt gi tr chn l ca mnh ab ngi ta khng quan tm n mi quan h v ni dung ca hai mnh a, b. Khng phn bit trng hp a c phi l nguyn nhn c b hay khng, m ch quan tm n tnh ng, sai ca chng. V d: "Nu mt tri quay quanh tri t th Vit Nam nm Chu u" mnh ng. V y hai mnh a = "mt tri quay quanh tri t" v b = "Vit Nam nm Chu u" u sai."Nu thng 12 c 31 ngy th mi nm c 13 thng" mnh sai.Ch :(mnh tng ng)Hai mnh a, b tng ng vi nhau hon ton khng c ngha l ni dung ca chng nh nhau, m n ch ni ln rng chng c cng gi tr chn l (cng ng hoc cng sai). V d: "Thng 12 c 31 ngy khi v ch khi tri t quay quanh mt tri" l mnh ng."12 gi tra hm nay Tun c mt H Ni nu v ch nu vo gi anh ang thnh ph H Ch Minh" l mnh sai."Hnh vung c mt gc t khi v ch khi 100 l s nguyn t" l mnh ng.-3-Gii bi ton bng suy lunV d:Ti Tiger Cup 98 c bn i lt vo vng bn kt: Vit Nam, Singapor, Thi Lan v Innxia. Trc khi thi u vng bn kt, ba bn Dung, Quang, Trung d on nh sau:Dung: Singapor nh, cn Thi Lan ba.Quang: Vit Nam nh, cn Thi Lan t.Trung: Singapor nht v Innxia nh.Kt qu, mi bn d on ng mt i v sai mt i. Hi mi i t gii my?Gii:K hiu cc mnh :d1, d2 l hai d on ca Dng.q1, q2 l hai d on ca Quang.t1, t2 l hai d on ca Trung.V Dung c mt d on ng v mt d on sai, nn c hai kh nng:Nu G(d1) = 1 th G(t1) = 0. Suy ra G(t2) = 1. iu ny v l v c hai i Singapor v Innxia u t gii nh.Nu G(d1) = 0 th G(d2) = 1. Suy ra G(q2) = 0 v G(q1) = 1. Suy ra G(t2) = 0 v G(t1) = 1.Vy Singapor nht, Vit Nam nh, Thi Lan ba cn Innxia t gii t.1. S v tTrong ton hc,s v t l s thc khng phi l s hu t, ngha l khng th biu din c di dng t s a/b , vi a, b l cc s nguyn.V d: S thp phn v hn c chu k thay i: 0.1010010001000010000010000001...S= 1,41421 35623 73095 04880 16887 24209 7...S pi = 3,14159 26535 89793 23846 26433 83279 50288 41971 69399 37510 58209 74944 59230 78164 06286 20899 86280 34825 34211 70679...S lgart t nhin e = 2,71828 18284 59045 23536...Nu nh mi s hu t u c biu din thp phn hoc hu hn (s thp phn hu hn, v d: 1/2=0,5) hoc v hn tun hon (s thp phn v hn tun hon, v d:1/11= 0.090909...) th s v t c biu bin thp phn v hn nhng khng tun hon.Cn bc hai ca tt c cc s nguynTa c th chng minh rng cn bc hai ca bt k s nguyn no cng phi hoc l s nguyn hoc l s v t.Ly s nguyn bt k r. Th d, r = 2.Trong h nh phn, 2 = 102Vy, nh trn, nu= m/n th, trong h nh phn:m2 = 102 n2 trong m, n l s nguynTrng hp n = 1 khng th xy ra, v ta bitkhng phi l s nguyn.Lp lun nh trn, v tri c s chn s 0 (trong h nh phn) cui, nhng v phi li c s l s 0 cui. Vy gi thitl s hu t phi sai.Vi s nguyn r bt k, cng chng minh nh trn trong h r-phn:m2 = 10r n2 trong m, n l s nguynNu n = 1 th m2 = 10r = r, vyl s nguyn.Cn nu n 1 th, nh trn, mt s bnh phng trong h r-phn phi c s chn s 0 (trong h r-phn) cui. Do trong ng thc ny v tri c s chn s 0 cui nhng v phi li c s l s 0 cui. Vykhng th l s hu t. 2. S chnh phngSchnhphnghaycn gils hnhvungl s nguyn c cn bc 2lmts nguyn, haynicchkhc, schnhphnglbnhphng(lythabc2)camt s nguyn khc.V d:4 = 2; 9 = 3; 1.000.000 = 1.000S chnh phng hin th din tch ca mt hnh vung c chiu di cnh bng s nguyn kia.-4-1 MNH 1.1 Xt xem cc cu sau, cu no l mnh , cu no l mnh cha bin?a) 7+x=3 b) 7+5=6 c) 4+x 1.8. Lp mnh PQ v xt tnh ng sai ca n, vi:a) P: 2;c) Nu A=900 th ABC l tam gic vung.-5-1.14. Dng k hiu hoc vit cc mnh sau:a) C mt s nguyn khng chia ht cho chnh n;b) Mi s thc cng vi 0 u bng chnh n;c) C mt s hu t nh hn nghch o ca n;d) Mi s t nhin u ln hn s i ca n.1.15. Pht biu bng li cc mnh sau v xt tnh ng sai ca chnga) x : x2 0 b) x : x20c) x : 2111xxx +d) x : 2111xxx +e) x : x2+x+1>0 f) x : x2+x+1>01.16.Lp mnh ph nh ca mi mnh sau v xt tnh ng sai ca na) x : x.1=x b) x : x . x =1c) n : n x2.b) x , |x| < 3 x< 3.c) x N, n2+1 khng chia ht cho 3.d) a , a2=2.1.20. Cc mnh sau y ng hay sai? Nu sai, hy sa li cho ng:A: 15 l s nguyn tB: a , 3a=7C: a , a231.21. Pht biu cc nh l sau, s dng khi nim "iu kin ":a) Trong mt phng, nu hai ng thng phn bit cng vung gc mt ng thng th ba th hai ng thng y song song nhau.b) Nu hai tam gic bng nhau th chng c din tch bng nhau.c) Nu mt s t nhin tn cng l chs 5 th chia ht cho 5.d) Nu a+b > 5 th mt trong hai s a v b phi dng.1.22. Pht biu cc nh l sau, s dng khi nim "iu kin cn":a) Nu hai tam gic bng nhau th chngc cc gc tmg mg bng nhau.b) Nu t gic T l mt hnh thoi th n c hai ng cho vung gc nhau.c) Nu mt s t nhin chia ht cho th n chia ht cho 3.d) Nu a=b tha2=b2.1.23. Pht biu nh l sau, s dng iu kin cn v Tam gic ABC l mt tam gic u khi v ch khi tam gic ABC l tam gic cn v c mt gc bng 6001.24. Hy sa li (nu cn) cc mnh sau y c mnh ng:a) t gic T l mt hnh vung, iu kin cn v l n c bn cnh bng nhau.b) tng hai s t nhin chiaht cho 7, iu kin cn v l mi s chia ht cho 7.c) ab>0, iu kin cn l c hai s a v b iu dng.-6-d) mt s nguyn dng chia ht cho 3, iu kin l n chia ht cho 9.1.25. Cc mnh sau y ng hay sai? Gii thch.a) Hai tam gic bng nhau khi v ch khi chng c din tch bng nhau.b) Hai tam gic bng nhau khi v ch khi chng ng dng.c) Mt tam gic l tam gic vung khi v ch khi c mt gc(trong) bng tng hai gc cn li.d) Mt tam gic l tam gic u khi v ch khi n c hai trung tuyn bng nhau v c mt gc bng 600.BI TP THM1. Xt ng (sai)ca mnh sau :a/ Hnh thoi l hnh bnh hnhb/ S 4 khng l nghim ca phng trnh : x2 5x + 4 = 0c/( 2>3 ) (3 < ) d/ (311 > 27) (42 < 0)e/ (5.12 > 4.6) (2 < 10) f)(1< 2 ) 7 l s nguyn t2. Ph nh cc mnh sau :a/ 1 < x < 3 b/ x 2 hay x 4c/ C mt ABC vung hoc cnd/ Mi s t nhin u khng chia ht cho 2 v 3e/ C t nht mt hc sinh lp 10A hc yu hay km.f/ x< 2 hay x=3.g/ x 0 hay x>1.h/ Ptx2 + 1 = 0 v nghim v pt x+3 =0 c nghim3. Xt ng (sai)mnh v ph nh cc mnh sau :a/ x R , x2 + 1 > 0 b/ x R , x2 3x + 2 = 0c/ n N , n2 + 2 chia ht cho 4 d/ n Q, 2n + 1 0e/ a Q , a2 > a f) x R , x2 +x chia ht cho 2.4.Dng bng ng (sai) chng minh:a) AB =B A b) A B A B c) A B A B d)( ) ( ) ( ) A B C A B A C B. SUY LUN TON HC5. Pht biu nh l sau di dng "iu kin "a/ Nu hai tam gic bng nhau th chng ng dng.b/ Hai ng thng phn bit cng song song vi ng thng th ba th chng song song vi nhau.c/ Nu a + b > 2 th a > 1 hay b > 1d/ Nu mt s t nhin c ch s tn cng l s 0 th n chia ht cho 5.e/ Nu a + b < 0 th t nht mt trong hai s phi m.6. Pht biu nh l sau di dng "iu kin cn"a/ Hnh ch nht c hai ng cho bng nhau.b/ Nu hai tam gic bng nhau th n c cc gc tng ng bng nhau.c/ Nu mt s t nhin chia ht cho 6 th n chia ht cho 3.d/ Nu a = b th a3 = b3.e/ Nu n2 l s chn th n l s chn.-7-7.Dng phng php phn chng, CMR :a/ Nu n2 l s chn th n l s chn.b/ Nu n2 l s chn th n l s chn.c/ Nu x2 + y2 = 0 th x = 0 v y = 0d/ Nu x = 1 hay y = 21 th x + 2y 2xy 1 = 0d/ Nu x 21 v y 21 th x + y + 2xy 21e/ Nu x.y chia ht cho 2 th x hay y chia ht cho 2.f) Nu d1// d2 v d1// d3 th d2 // d3.8. Chng minh vi mi s nguyn dng n, ta c:a) 1 + 3 + 5 + 7 + .. . . . . . . . +(2n 1) = n2 b) 2 + 4 + 6 + 8 + . . . . . . . . . . + (2n) = n(n +1)c) 1 + 2 + 3 + 4 +. . . . . . . . . + n = 2) 1 n ( n + a) 1.2 + 2.3 + 3.4 + . . . . .+ n.(n + 1) = 3) 2 n )( 1 n ( n + +b) 1 nn) 1 n .( n1.........4 . 313 . 212 . 11+++ + + +c) 1 n 2n) 1 n 2 ).( 1 n 2 (1.........7 . 515 . 313 . 11++ + + + +d) 12 + 22 + 32 + . . . . . . . . . . + n2 = 6) 1 n 2 )( 1 n ( n + +e) 13 + 23 + 33 +. . . . . .+ n3 = 4) 1 n ( n2 2+f) 2 1 + 22 + 23 +. . . . .+ 2 n = 2(2 n 1)g) 31 + 32 + 33 +. . . .+ 3 n = 23( 3 n 1 )h) n 3 +2n chia ht cho 3 i) n3 +11n chia ht cho 6 j) n3 +5n chia ht cho 6k) 3 2n + 63 ht 72l) 3 2n + 1 + 2 n + 2chia ht cho 7m) 6 2n+ 3 n + 2 + 3 nchia ht cho 11n) 3 2n 2 nchia ht cho 7o) 4 n + 15.n 1 chia ht cho 91 MNH 1.3. a) PQ: Nu gc A bng 900 th BC2=AB2+AC2ng QP: Nu BC2=AB2+AC2 th gc A bng 900 ngb) PQ: A B th tam gic ABC cnngQP: Nu tam gic ABC cn th A B sai (v c th A C 1.4. a) x : x2=1; C mt s thc m bnh phng ca n bng 1sai x : x21; Vi mi s thc, bnh phng ca n u khc 1b) x :x2+x+20; Vi mi s thc u c x2+x+20 ngx :x2+x+2=01.5. a) ng. P: 13 23 2+ -8-b) Sai.P: ( )22 8 8 c) ng v ( )23 12 +=27 l s hu t. P: ( )23 12 +l s v td) Sai. P: x=2khngl nghim ca phng trnh 2402xx1.8. Lp mnh PQ v xt tnh ng sai ca n, vi:a) Nu 2; ng v m o ngc) Nu A=900 th ABC l tam gic vung. ng v m o sai (vung ti B hoc C)1.14. a) n : n khng chia ht cho n b) x : x+0=0c) x : xn1.15. Pht biu bng li cc mnh sau v xt tnh ng sai ca chnga) Bnh phng mi s thc u nh hn hoc bng 1saib) C mt s thc m bnh phng ca n nh hn hoc bng 0ngc) Vi mi s thc , sao cho2111xxx +Said) C s thc, sao cho 2111xxx +nge) Vi mi s thc x, sao cho x2+x+1>0ngf) C mt s thc x, sao cho x2+x+1>0ng1.16. a) x : x.1x saib) x : x . x 1ngc) n : nn2 ng1.17.a) C t nht mthnh vung khng phi l hnh thoisaib) Mi tam gic cnl tam gic usai1.18.Xt xem cc mnh sau y ng hay sai v lp mnh ph nh ca mi mnh :a)x , 4x2-1= 0sai; m ph x , 4x2-10b)n , n2+1 chia ht cho 4Sai v Nu n l s t nhin chn : n =2k (kN)n2+1 = 4k2+1 khng chia ht cho 4Nu n l s t nhin le : n = 2k+1 (kN)n2+1 = 4(k2+k)+2 khng chia ht cho 4M ph nh n , n2+1 khng chia ht cho 4c)x , (x-1)2 x-1. Sai khi x=0m ph nh x ,(x-1)2 =x-11.19. a) ng, v d x=1/10b) sai, v khi x f(x2)Bng bin thin: l bng tng kt chiu bin thin ca hm s (xem SGK)III. Tnh chn l ca hm s+ f gi l chn trn Dnu xD x Dvf(x) = f(x), th nhn Oy lm trc i xng.+ f gi l l trn Dnu xD x Dv f(x) = f(x), thnhnO lm tm i xng.(Ban CB n III)* Tnh tin th song song vi trc ta OxyCho (G) l th ca y = f(x) v p;q > 0; ta c Tnh tin (G) ln trn q n v th c th y = f(x) + qTnh tin (G) xung di q n v th c th y = f(x) qTnh tin (G) sang tri p n v th c th y = f(x+ p)Tnh tin (G) sang phi p n v th c th y = f(x p)i xng qua trc honh th x khng i y= -yi xng qua trc tung th y khng ix= -x* Tnh tin im A(x;y) song song vi trc ta Oxy :+ Ln trn q n v cA1(x ; y+q)+ Xung di q n v cA1(x ; yq)+ Sang tri p n v c A1(xp ; y)+ Sang phi p n v c A1(x+p ; y)-20-CC DNG BI TPI. Tm tp xc nh ca hm s*Phng php + tm tp xc nh D ca hm sy = f(x)ta tm iu kin f(x) c ngha,tc l: D = {x | f(x) } + Cho u(x), v(x) l cc a thc theo x , khi ta xt mt s trng hp sau :a)Min xc nh ca hm s dng ng thc : y=u(x) ; y = u(x)+v(x) ; y=| u(x) | ; y =| ) ( | x u l D =(khng cha cn bc chn, khng c phn s, ch c cn bc l,)b)Min xc nh hm sy = ) () (x vx ul D = { x | v(x) 0 }c)Min xc nh hm s y = ) (x u lD = { x | u(x)0 } d) Min xc nh hm s y = ) () (x vx u lD = { x| u(x) > 0 }e) Min xc nh hm s y = ) ( ) ( x v x u +lD= {x | u(x)0 } {x | v(x)0 } tc l nghim ca h'0 ) (0 ) (x vx u V D : Tm tp xc nh ca cc hm s sauII. Xt s bin thin ca hm s* Phng php + Tm tp xc nh Dca hm s y = f(x). + Vit Dv dng hp ca nhiu khong xc nh ( nu c ). + Xt s bin thin ca hm s trn tng khong xc nh K= (a;b) nh sau: . Gi sx1,x2 K,x1 < x2 . Tnh f(x2) - f(x1) . Lp t s T = 1 21 2) ( ) (x xx f x f NuT > 0 th hm sy = f(x) ng bin trn (a;b) NuT < 0 th hm sy = f(x) nghch bin trn (a;b). V D:III.Xt tnh chn l cahm s * Phng php + Tm tp xc nh Dca hm s y =f(x) + Chng minh Dl tp i xng, tc l :xD x D + Tnh f(-x), khi . Nu f(-x) = f(x)vixDth y =f(x) l hm s chn. Nu f(-x) = -f(x) vixDth y = f(x) l hm s l. . Nu c mt x0 Dsao f(-x0) f(x0) & f(-x0) -f(x0) th hm s y = f(x) khng chn v khng l. V D:IV. Tnh tin th song song trc ta Cho (G) l th ca y = f(x) v p;q > 0; ta c Tnh tin (G) ln trn q n v th c th y = f(x) + qTnh tin (G) xung di q n v th c th y = f(x) qTnh tin (G) sang tri p n v th c th y = f(x+ p)Tnh tin (G) sang phi p n v th c th y = f(x p)-21-BI TP 1-C21.1. Tm tp xc nh ca cc hm s saua) y= 3x3x+2 b) 3 12 2xyx +c) y= 3 2 x d) y= 2 1 1 x x + e) y=22 12 1xx x+ +f) y=11 xx + +g) y=21 x +h) 214 5yx x+ +1.2. Cho hm s y=1- x neu x 0xneu x>0 '. Tnh cc gi tr ca hm s ti x=3; x=0; x=11.3. Cho hm s y=22 3012 0xkhi xxx x khi x' + >Tnh gi tr ca hm s ti x=5; x=2; x= 21.4. Cho hm s y=g(x)3 87 2vix 4/33.nh m hm s xc nh vi mi x dng a/ 1 4 y x m x m + b/ 2x my x mx m + ++4. Xt s bin thin ca cc hm s trn khong ch ra :a/ y = x2 4x (-, 2) ; (2, +)b/ y = 2x2 + 4x + 1 (-, 1) ; (1, +)c/ y = 1 x4+(1, +)d/ y = x 32 (3, +)e/ y = 1 xx 3 (, 1)f/ y = 1 x 6. Xc nh tnh chn, l ca hm s :a/ y = 4x3 + 3x b/ y = x4 3x2 1c/ y = 3 x12+d/ y = 2x 3 1+e/ y = |1 x| + /1 + x| f/ y = |x + 2| |x 2|g/ y = |x + 1| |x 1| h/ y =x 1+x 1+i/ y = | x|5.x3k/x x2+x xy|2 +|+|2 || ||2 |l/ y = ' +1 11 1 01 122x ; xx ;x ; x m) y = ' 11 1 0122x ; xx ;x ; x-29-yyxxOODDCCBBAA444422yyxxO2 HM S y= ax + b1. Hm s bc nhtHm sdngy = ax + b , a;b v a 0. H s gc l aTp xc nh: D = Chiu bin thin:a > 0 hm s ng bin trn a < 0 hm s nghch bin trn Bng bin thin: th hm s: l mt ng thng. th khng song song v trng vi cc trc ta ,ct trc tung ti im (0;b) v ct trc honh ti (-b/a;0).2. * Cho hai ng thng (d):y= ax+b v (d)= ax+b, ta c:(d) song song (d) a=a v bb(d) trng (d) a=a v b=b(d) ct (d) aa.(d)(d) a.a= 12. Hm s hng y=b ng thng y= b l ng thng song song hoc trng trc Ox v ct Oy ti im c ta (0;b).ng thng x= a l ng thng song song hoc trng trc Oyv ct Ox ti im c ta (a;0)3.Hm s bc nht trn tng khong, hm s y= |ax+b|Mun v th hm s b ax y + ta lm nh sau:+ V hai ng thng y = ax + b, y = - ax b+ Xa i hai phn ng thng nm pha di trc honhV d 1: Kho st v v th hm s y= |x| (Xem SGK tr.42)V d 2: Xt hm s y=f(x)=' < + < +5 x 4 n e u4 x 2 n e ux 0 n e u6 24212 1xxx th (hnh)V d 3 : Xt hm s y=|2x-4|Hm s cho c th vit li nh sau : y='< + 2 x n e u2 x n e u4 24 2xx th (hnh)V d 4: Tm hm s bc nht y=f(x) bit th ca n i qua 2 imA(0 ; 4) , B (-1;2).V th v lp bng bin thin ca hm s ( ) ( ) y g x f x .-30-GiiHm s bc nht c dng , 0 y ax b a + . th hm s qua im A , B 4 22 4b aa b b ' ' + V th hm( ) 2 4 g x x + , ta v th hai hm s

' < + 2 x neu2 x neu 4 24 2xxy trn cng 1 h trc ta , ri b i phn pha trn trc Ox.V th hm( ) 2 4 g x x +xyo-2-4-4 Bng bin thin.g(x)-2 x+0BI TP 2-C22.1. V th cc hm s saua) y= 2x+1 b) y= 3 c) y= 273x e) y= 23 xf) y=35 x 2.2. V th cc hm s sau:a) y=|x|+2x b) y= |3x2|c) 21 2vi x>2vixxy+ 'd) 2 1 1112vi xvi x0 th hai phng trnh x2=1 v x=1 tng ng nhau.2. Php bin i tng ng:php bin i mt phng trnh xc nh trn D thnh mtphng trnh tng nggi l php bin i tng ng trn D.(ta dng du "" ch s tng ng ca cc phng trnh)V d: 2x-5=0 3x215=0 -39-* Cc php bin i tng ng ca phng trnh:nh l : Cho phng trnhf(x) = g(x)c tp xc nh D. nu h(x) xc nh trn D th phng trnh:) ( ) ( ) ( ) ( ) ( ) ( x h x g x h x f x g x f + + D x h(x) x h x g x h x f x g x f moi vi neu 0 ) ( ) ( ) ( ) ( ) ( ) (H qu : Nu chuyn mt biu thc t mt v ca mt phng trnh sang v kia v i du ca n th ta c phng trnh mi tng ng vi phng trnh cho.* Ch : Nu 2 v phng trnh lun cng du th khi bnh phng hai v ca n, ta c phng trnh tng ng.V d 1: 3. Phng trnh h qua) nh ngha:f1(x)=g1(x) gi l phng trnh h qu ca phng trnh f(x)=g(x) nu tp nghim ca n cha tp nghim ca phng trnh f(x)=g(x). Khi ta vit: f(x)=g(x)f1(x)=g1(x)b) Php bin i cho phng trnh h qu :Khi bnh phng hai v ca mt phng trnh ta i n phng trnh h qu.* Ch : Phng trnh h qu c th c thm nghim khng phi l nghim ca phng trnh ban u. Ta gi l nghim ngoi lai. Khi ta phi th li cc nghim loi b cc nghim ngoi lai.V d 1: Gii phng trnh 22 2 3 72 2 4x x xx x x+ ++ + (1)iu kin pt(1)l x2 v x2(1) (x+2)2+(x2)2= 3x+7Hoc: Vi iu kin x2 v x2 th (1)(x+2)2+(x2)2= 3x+7 (???)V d 2: a) |x2|=x+1 (x2)2=(x+1)2b)1 x =x x1= x2.V d 3: Gii phng trnh2 x x (3)Gii iu kin x 0. Bnh phng hai v phng trnh (3) x24x+4 = xx25x+4=0(3')Phng trnh (3') c nghim x=1 hoc x=4Th li vo phng trnh (3), ta thy x=1 khng phi l nghim ca (3) v x=4 l nghim. Vy pt(3) c ngim duy nht x=4.BI TP P DNG1/ Tm iu kin ca cc phng trnh a) 2234xxx b) 412xxx+ c) 12 1 xx+ d) 2223 12 1xx xx+ + ++e) 21 3xx x +f) 22 314xxx+ +p sa) x 3, x 2 b) Khng c gi tr x tha c) x1/2 v x0 d) x Re) x>1f) x1 v x22/ Chng t cc phng trnh sau v nghim-40-a) 3 132xxx+ +b)4 3 4 x x x + 3)Gii cc phng trnh saua)1 3 1 x x x + + + + b)5 2 5 x x x + c) 2 1 33 3x xx x+ + d) 22 81 1xx x+ +S: a) x=3 b) V nghim c) V nghim d) x=24) Gii cc phng trnh saua)1 1 2 x x x + + + + b)3 3 3 x x x +c) 22 3 4 x x x + d) 21 4 1 x x x + + e) 23 1 41 1xx x+ f) 23 444x xxx+ + ++g) 23 23 23 2x xxx h) 24 32 31 1xxx x ++ + p s: a) x=2 b) x=3 c) VNod) x=2e) VNof) x=0 v x=2 g) x=4/3 h) x=25) Cho phng trnh (x+1)2 =0 (1) vax2(2a+1)x+a=0 (2)Tm a (1) tng ng (2)HDGi s (1)(2) th x= 1 ca (1) l nghim ca 2. Th x=1 v (2) ta tm c a=1/4.Khi a=1/4 th vo (2) (x+1)2=0Vy (1) (2)6) Tm m cc cp pt sau tng nga) x+2=0 v 3 1 03mxmx+ +b) x29=0 v 2x2+(m5)x3(m+1)=0c) 3x2=0 v (m+3)xm+4d) x+2=0 v m(x2+3x+2)+ m2x+2=0p s: a) m=1 b) m=5 c) m=18 d) m=1BI TP(i cng v phng trnh)1/ Tm iu kin xc nh ca mi phng trnh sau ri suy ra tp nghim ca n) )3 2 2 63) 3 ) 13a x x b x x xxc x x d x x xx + + + 2/ Gii cc phng trnh sau ) 1 2 1 ) 1 0,5 13 2) )2 5 5 2 5 5a x x x b x x xx xc dx x x x+ + + + 3/ Gii cc phng trnh sau11 211)+xxxx a23 221)+xxxx b0 3 ) 2 3 ( )2 + x x x c 0 1 ) 2 ( )2 + x x x d-41-4/ Gii cc phng trnh sau bng cch bnh phng hai v1 2 | 2 | ) 2 | 1 | 2 )3 1 ) 2 9 3 ) + x x d x x cx x b x x a5/ Tm nghim nguyn ca mi phng trnh sau bng cch xt iu kina) 4 x - 2 =x- x b) 3 2 x +=2 x + 2 26/ Gii cc phng trnh sau :a/ 1 x = x 1b/ x +3 x = 3 +3 x c/ 4 x + 1 = x 4 d/ x + x = x 2e/ 2 x2 x = 2 x1f/ 1 x3 = 1 x2 x+g/ 3 x1 x = x 327/ Gii cc phng trnh sau :a/ x + 2 x1 = 2 x1 xb/ 1 x (x2 x 6) = 0c/ 1 x2 x x2+ + = 0 d/ 1 + 3 x1 = 3 xx 2 7e/ 2 x9 x2+ = 2 x3 x++8/ Gii cc phng trnh :a/ x 1= x + 2 b/ x + 2= x 3 c/ 2 x 3= x + 1d/ x 3= 3x 1 e/ xx 1 = xx 1f/ 2 xx = 2 xxg/ x1 x = x1 x h/ 3 x2 x = 3 xx 2BI TP THMBi 1: Gii cc phng trnh saua) x = x b) 3 x = 3 x +1 c) x+ 2 x = 2+ 2 - xd) x+ 2 x = 1+ 2 xe)131 x xx f)111 x xx.Bi 2: gii cc phng trnh sau a)11 211+xxxx b)23 221+xxxx c) 3 x (x2-3x+2) = 0 d) 1 + x (x2-x-2) = 0 e)2212 xx xx f)113142+ ++++xxxxx .Bi 3: Gii cc phng trnh sau a)1 2 + x x b)2 1 + x x c)2 1 2 + x x d)1 2 2 x x e) 3 9 2 x x f) 1 3 x x -42-Bi 4: Gii cc phng trnh saua)1 1 xxxx b)1212xxxx c)xxxx 2 2 d)2121xxxx .-43-2 PHNG TRNH QUY V PHNG TRNH BC NHT,BC HAII. Phng trnh bc nht, phng trnh bc hai1. Phng trnh bc nht Gii v bin lun phng trnh dng ax+b = 0 a 0: Phng trnh c nghim duy nht x= ba a = 0 v b 0: Phng trnh v nghim a = 0 v b=0: Phng trnh nghim ng vi mi x(v s nghim)* Ch :+ Trc khi gii v bin lun phng trnh bc nht ta phi a phngtrnh v dng ax+b = 0 .+ khi bin lun a=0 th thay gi tr m va tm c vo b . + Khi a0 th phng trnh ax+b = 0 mi c gi l phng trnh bc nht mt n.V d1. Gii v bin lun phng trnh : m(x- m ) = x + m -2 (1)Gii Phng trnh (1) (m - 1)x = m2 + m 2 (1a)Ta xt cc trng hp sau y :+ Khi (m-1) 0 m 1nn phng trnh (1a) c nghim duy nht x = 221m mm+ = m 2;nn pt(1) c nghim duy nht+) Khi (m 1) = 0 m = 1 . phng trnh (1a) tr thnh 0x = 0; phng trnh nghim ng vi mi x R; nn pt(1) ng vi mi x R.Kt lun :m 1 : nghim l x= m-2 (Tp nghim l S = {m - 2}) m = 1 : ng x R(Tp nghim l S = R)V d 2: Gii v bin lun phng trnh: m(x-1) = 2x+1 (2)Gii Ta c (2) mx-m = 2x+1 (m-2)x = m+1(2a)(c dng ax+b =0)Bin lun:+ nu m-20 m2 th (2a) c nghim duy nht21+mmx + nu m-2= 0m = 2 th (2a)tr thnh 0x=3; pt ny v nghim, nn (2) v nghim.Kt lun: m2 th (2) c nghim 21+mmx m=2 th (2) v nghim. V d 3: Gii v bin lun phng trnh m2x+2 = 2m-2 (3)Gii Ta c: (3) m2x-x = 2m-2 (m2-1)x = 2(m-1)(3a)Bin lun: + Nu m2-10 m1 t th (3a) c nghim duy nht 121) 1 ( 22+m mmx ; nn (3) c nghim duy nht. + Nu m2-1=0 m=1 t-44-- vi m=1 :(3a) c dng 0x= 0, (3a) ng vi mi xR (phng trnh c v s nghim), nn (3) c v s nghim.- vi m=-1: (3a) c dng 0x=-4; (3a)v nghim, nn (3) v nghim.Kt lun: + m1 v m -1 th (3) c nghim duy nht 12+mx + m =1 th (3) c v s nghim+ m= -1 th (3) v nghim.V d 4: Gii v bin lun phng trnh sau theo tham s m: (*) 113+ xm mx GiiVi x-1 th (*) mx-m-3 = x+1 (m-1)x = m+4 (**) Bin lun (**) vi x-1 + Nu m 1 th (**) c nghim 23114114 + + mmmmmx + Nu m=1: (**) 0x=4, v nghim Kt lun : m1 vm23 th (*) c nghim x=14+mm

231m th (*) v nghim V d 5:gii v bin lun phng trnh theo tham s m: 2 3 1 + + m x mx (1) GiiTa c (1)

+ + + +(3) 2 m - -3x 1 mx(2) 2 3 1 m x mx + gii v bin lun (2)(2) (m-3)x= m-3 . nu m3 th (2) c nghm x=1 . nu m=3 th(2)0x = 0 =>(2) c v s nghim+ gii v bin lun (3) (3)(m-3)x=-m+3 . nu m-3 th (3) c nghim x=31++ mm . nu m = -3 th (3) 0x=4, v nghim Kt lun: - vi m3 v m-3 : (1) c hai nghim x1=1 v x2 =31++ mm - vi m=3: (1) c v s nghim - vi m=-3:(1) c nghim x=1(v tha phng trnh (2) )2. Phng trnh bc hai (nhc li cch gii phng trnh bc hai)Gii v bin lun phng trnh dngax2+bx+c = 0 a= 0 :Tr v gii v bin lun phng trnh bx + c = 0 a 0 . Lp = b2 4ac (hoc =b2-ac) Nu > 0: phng trnh c hai nghim phn bit -45-x = 2ba vx = 2ba + Nu = 0 : phng trnh c nghim kp : x = 2ba Nu < 0 : phng trnh v nghimV d 1: Gii v bin lun phng trnh mx2-2(m+1)x+m+1 = 0 GiiPhng trnh cho c dng phng trnh hc. Bin lun: . Nu m = 0 ( thay m = 0 vo phng trnh ta c -2x+1= 0 => x=21

. Nu m 0 , tnh' = m+1, khi : + nu ' < 0 m < -1 pt v nghim + nu' = 0 m = -1 pt trnh c nghim kp x1=x2 = 0 + nu' > 0 m > -1 pt c hai nghim phn bit x1,2 = mm m 1 1 + t + * Kt lun: V d 2: nh m phng trnh mx2-2(m-2)x+m-3 = 0 c nghim3. nh l VitNu phng trnh bc hai ax2+bx+c = 0 (a0) c hai nghim x1, x2 th tng (S) v tch (P) ca hai nghim l: S = x1+x2 = abP = x1.x1 = acNgc li, nu hai s u, v c S=u+v; P=u.v th u, v l nghim ca phng trnh x2-Sx+P = 0. V d 1: tm hai s bit S =19 , P = 84 GiiHai s cn tm l nghim ca phng trnh bc hai x2-19x+84 = 0 ,pt ny c hai nghim

12721xxhoc

12721xx vy hai s cn tm l 7 v 12. * Ch : iu kin phng trnhx2-Sx+p =0 c nghim l S24P . y cng l iu kin tn ti hai s c tng l S, tch P. * ng dngP S x x x x x x 2 2 ) (22 122 12221 + +

PSx x +2 11 1 PS S x x x x x x x x 3 ) ( 3 ) (32 1 2 132 13231 + + +4 41 2x x + ( )22 2 2 21 2 1 22 x x x x + =(S22P)22P2V d 1: Cho phng trnh x24x+m1= 0. Xc nh m phng trnh c hai nghim2 21 2x x +=10.iu kin pt c nghim '0 5m0 m5S22P = 10 m =4.V d 2: Xc nh m phng trnhx2-4x+m-1= 0 c hai nghim x1, x2 tha x thc 403231 + x xGii Phng trnh c nghim ' 0 5 m Theo gi thit403231 + x x S3-3PS=40 64-12(m-1)=40 m= 4 (nhn) -46-* Xt du cc nghim phng trnh bc hai Gi s phng trnh bc hai c hai nghim x1,x2 th: x1< 0 < x2 P < 0 (hai nghim tri du) x1x2 < 0 '< >000SP ( hai cng m) 0 < x1x2 '> >000SP(hai cng dng)V du: cho phng trnh x2+5x+3m-1 = 0 (1) a) nh m e phng trnh co hai nghiem trai dau.b) nh m e phng trnh co hai nghiem am phan biet. Giai a) pt(1) co hai nghem trai dau P < 0 0 1 3 0 < < mac m < 31

b) e phng trnh co hai nghiem am phan biet ' >000SP'< > + > 0 50 4 1 2 2 50 1 3mm '1 22 931mm 311229< b nu ab > 0. Khi ta cng k hiu b b a-b > 0 (bab v c>d (hoc a b a+c > b+c(cng 2 v bt ng thc cng 1 s)a > b+ c ac > b (chuyn v) 3)a > b ac bc > > < + '>>5)bd acd cb a> '> >> >006) Vi n nguyn dng: a > b a2n+1 > b2n+1a > b>0 a2n > b2n7) Nu b>0 th a>b a b > ; a>b 3 3a b > 8) c ac bb a> '>>(bc cu)9) a > b < >> b > 0 an > bn( n+N) 11)a > b > 0 n nb a >( n+N)Ch : Khng c quy tc chia hai v bt ng thc cng chiu -67-PHNG PHP CHNG MINH BT NG THCPhng php chung: Mt s hng ng thc:(at b)2= a2 t2ab +b2(a+b+c)2= a2+b2+c2+2ab+2ac+2bc(at b)3= a3 t3a2b+3ab2 tb3a2 b2 = (ab)(a+b)a3b3= (ab)(a2 +ab +b2)a3b3= (a+b)(a2 ab +b2)V d: Chng minh rnga) Nu a,b0 th a+bab 2b) Chng minh a2+b2-ab 0. Khi no th ng thc xy ra.Giia) Cch 1: ta c a+bab 2 a+b- ab 2 0 ( b a )2 0ng vi mi a,b0. Du '=' xy ra khi a = bCch 2: ta bit( b a )200 , b a a+b- ab 20 a+bab 2pcm. b) Ta c:a2+b2-ab =ab b b a + +2 2 24341 = (a-2)2b +R b a, 0432 b du '=' xy ra '' 00043022babba pcm4. Bt ng thc Csi a/ nh l: Nu a0, b0 th abb a+2haya+bab 2 Du '=' xy ra a=b b/ Cc h qu: b.1. N a0,b0 c a+b=const (hng s) tha.b max a = b b.2. Nu a0,b0c a.b = const th a + b l min a = b b.3.Nu a1, a2, a3,..,an0 th: nnna a a ana a a... ....3 . 2 12 1+ + +b.4. 12 aa+ , a > 0 * ngha hnh hc: -68-+ Trong tt c cc hnh ch nht c cng chu vi, hnh vung c din tch ln nht.+ Trong tt c cc hnh ch nht c cng din tch, hnh vung c chu vi nh nht.c. V d:V d 1: cho hai s a, b> 0. Chng minh rng 2 +abba Giip dng bt ng thc Csi cho hai s dng 0 , >abba ,ta c: 2 2 . 2 + +abbaabbaabba => pcm. V d 2: Chng minh rng vi a,b>0 th (a+b)(ab+1)4abGii Ap dng bt ng thc Csi cho hai s dng a,b>0 ta c: a+b2 ab(1) Ap dng bt ng thc Csi cho hai s dng ab,1>0 ta c: ab + 12ab(2)Nhn (1) vi (2) ta c: (a+b)(ab+1)4ab=> pcm 5. Bt ng thc cha gi tr tuyt i nh ngha: |x| =' pcm6. Bt ng thc Bunhiacopxki Cho 4 s thc a, b, c, d bt k th:(ab+cd)2 (a2+c2)(b2+d2) ) )( (2 2 2 2d b c a cd ab + + + Chng minh: Ta c (ab+cd)2 (a2+c2)(b2+d2) a2b2+c2d2+2abcd a2b2+a2d2+b2c2+c2d2 a2d2+b2c2-2abcd 0 (ad-bc)20 ng R d c b a , , ,=> pcmV d 1: cho x2+y2=1,chng minh rng2 2 + y x GiiAp dng bt ng Bunhiacopxki cho bn s a = 1, b = x, c = 1, d = y ta c:(1.x+1.y)2(12+12)(x2+y2) (x+y)22 2 2 + y x => pcm. V d 2: Cho x+2y = 2 , chng minh rng x2+y254-69- Gii Ap dng bt ng thc Bunhiacopxki cho bn s a = 1, b = x, c = 2, d = yBI TP P DNG1/ Vi mi s thc x, y, z . Chng minh rng:2 2 22 y xyz x z +HD: a v hng ng thc2/ Chng minh rng:11 1 , a 1 a aa< + Gii( )222 22221 11 1 1 11 1 1( 1) ( 1) 2 a 1 2 a 1 2 . V 2 0 nn1 14(a 1) 2 0aa a a aa aa a a aa a aaa _< + < + , < + + < > _ < > , ungVy 11 1 , a 1 a aa< + pcm3/ Tm Gi tr nh nht ca hm s y= 1 11 x x+ vi 00 nn p dng bt C-si cho hai s dngta c: y= 1x+11 x 1 1 12 . 21(1 )x xx xm + + (1 ) 1 1(1 )2 (1 )(1 )2x xx xx xx xvy y= 1x+11 x + 1 1 1 12 . 2 2 41 (1 )(1 )2x x x xx xy= 1x+11 x4. Du "=" xy ra '1 1112(0;1)xx xxVy gi tr nh nht ca hm s y= 1x+11 x bng 4 khi x =12BI TP1/ Cho a, b, c, d l nhng s dng; x, y, z l nhng s thc ty . Chng minh rng:a) 4 4 3 3y y x x y x + +Gii4 3 4 3 3 33 3 3 3222 2 2 2( ) 0 ( ) y ( ) 0( ) y ( ) 0 ( )( y ) 03( ) ( y ) 0 ( ) 02 4a x x y y y x x x y y xx x y x y x y xy yx y x xy x y x ng + + 1 _ + + + + 1 , 1 ]Vy 4 4 3 3y y x x y x + + pcmb) 2 2 24y 3 14 2 12 6 x z x y z + + + > + +-70-Gii2 2 22 2 2( ) 2 1 4y 2.2 .3 9 3 2. 3. . 3 3 1 0( 1) (2 3) ( 3. 3) 1 0 b x x y z zx y z + + + + + + > + + + > ungVy2 2 24y 3 14 2 12 6 x z x y z + + + > + +pcmc)* a ba bb a+ +Gii( ) ( )3 32( )( )( ) ( )( )( ) ( ) 0 ( )( ) 0 ( )( 2 ) 0( )( ) 0 a ba a b bc a b a bb a b aa b a a b b b a a ba b a a b b b a a ba b a a b b b aa b a a b b a b a b++ + + + + + + + + + + + + + pcmd) 1 1 4a b a b+ +Giip dng bt C-si cho hai s dng a, b: 2 a b ab + (1)p dng bt C-si cho hai s dng 1 1 1 1 1, : 2a b a b ab+ (2)Ly (1) nhn (2) ta c: 1 1 1 1 4( )( ) 4 a ba b a b a b+ + + +. pcme)* 44a b c dabcd+ + +(bt C-si cho 4 s)Gii4422( ) 2.2 424a b aba b c d ab cd ab cd abcdc d cda b c dabcd+ + + + + )+ + + + f) 1 1 1 1 16a b c d a b c d+ + + + + +Giip dng bt C-si cho 4 s dng a, b, c, d ta c:44 a b c d abcd + + + (1)p dng bt C-si cho 4 s dng1 1 1 1, , ,a b c d ta c;41 1 1 1 14a b c d abcd+ + + (2)Nhn (1) vi (2) ta c: 1 1 1 1( )( ) 16 a b c da b c d+ + + + + + Vy 1 1 1 1 16a b c d a b c d+ + + + + +g) 21a 2 b ab+ p dng bt C-si cho 2 s dng a2b, 1/bh) ( )( )( ) 8 a b b c c a abc + + + p dng bt C-si cho a, b v b, c v c, a.-71- i) ( )22 2( ) a b a b ab + +Khai trin hng ng thc ri p dng bt C-si cho ( ) a b +v2 abj) 1 1 1 9a b c a b c+ + + +Giip dng bt C-si cho 3 s dng a, b, c ta c:33 a b c abcd + + (1)p dng bt C-si cho 3 s dng1 1 1, ,a b c ta c;31 1 1 13a b c abc+ + (2)Nhn (1) vi (2) ta c: 1 1 1( )( ) 9 a b ca b c+ + + + Vy 1 1 1 9a b c a b c+ + + +2/ Chng minh cc bt ng thc saua) Vi x>3. Chng minh 423xx++HD: 4 2 3 x x + + p dng bt C-si cho 1 v x+3b) Vi 2 2y14 9x+ . Chng minh |x.y|3HD: p dng bt C-si cho 24x,2y9c)* Vi a, b, c 0 v a+b+c=1. Chng minh: b+c 16abcHD:b+c 2 bc (b+c)2 4bc (1) a+(b+c) 2 ( ) a b c + 14a(b+c)(2)ly (1)x(2) ta cpcmd) Cho a, b, c, d 0. Chng minh: (abc+2)(bc+2)(a+d)(d+1) 32abcdHD: p dng bt C-sicho: abc v 2; bc v 2; a v d; d v 1e) Cho a,b,c >0. CMR :8 ) 1 )( 1 )( 1 ( + + +accbbaHD: p dng bt C-sicho 1, ;1, ;1,a b cb c af) Vi a,b,c,d khng m. CMR : (a+b)(b+c)(c+d)(d+a) 16abcd.HD: g) Cho a,b,c > 0. CMR :2bca abc+ HD: h) Cho a,b,c > 0. CMR : (a+b+c)(c b a1 1 1+ + ) 9HD: k) Cho a,b > 0. CMR : (a+b)(1 1a b+ ) 4HD: l) Cho a,b,c > 0. CMR : 422a bcabc+HD: 422 22 2a bc aab bc abc c+ + -72-m) Cho a,b,c > 0 v a+b+c =1.CMR :64 )11 )(11 )(11 ( + + +c b aHD: n) Cho a > 1 . CMR : 21aa HD: bnh phn 2 vo) Cho a,b,c >0 . CMR : 1 1 1 1 1 1a b cab bc ac+ + + +3/ Chng minh bt ng thca) Chng minh rng nu a > b > 0 th 1 1b a>b) 2 2 2a ,a,b,c b c ab bc ca + + + + . Khi no du "=" (ng thc) xy ra?c) 2 2a 0, , b ab a b + + . Khi no du "=" (ng thc) xy ra.?d) (a+b+c)23(a2+b2+c2) vi mi a,b,c . e)a2b+ab2a3+b3 , vi a, b dng. ng thc xy xy ra khi no ?4/ Cho hm s f(x) = (x+3)(5-x) vi 5 3 x . Xc nh x sao cho f(x) t gi tr ln nht?5/Tm gi tr nh nht ca cc hm s sau a) f(x)= 0 x vix3x > + b) f(x)= 11+xxvi x > 1 2*/ Tm gi tr nh nht ca hm s y= 4 91 x x+ vi 0 0 aR3. a2 + b2 + 4 ab + 2(a +b) , a, bR4. a2+ b2 + c2 + d2 + e2 a(b +c + d + e) , a, b, c, d, eR5.2411 2aa Ra + , . Suy ra 2 24 411 1a ba b+ + + , a, bR6.22 2 23 3a b c a b c + + + + _ , , a, b, cR 7. a3 + b3 ab(a+b) , a, b 08.a3b + ab3 a4 + b4 , a, bR 9. a4 + 16 2a3 + 8a, aR10. ( )( ) a b c d ac bd + + +, a, b, c, d > 011.a ba bb a+ + ,a, b > 0 12.2 232a ab b a b + + +, a, bR 13.11 1 a aa + < + ,a 114.2 2 2a b ca b cb c a+ + + +, a, b, c > 015. a4 + 2a3 +3a2 -12a +19 > 0 , aR16. x8 x5 + x2 x + 1 > 0 , xR.Hd: BT 5 32 3( 1) ( 1) 1 0 (1 ) (1 )x x x xx x x + + > + + 8neu x1xneu x 0, b > 0, c > 0 . CMR: i.Nu 1a a cb b c+< >+a th b b/Cho a > 0, b > 0, c > 0 . CMR: 1 2a b ca b b c c a< + + 03.Cho a + b = 1. CMR: a2 + b2 124.Cho x + y + z = 1. CMR: 2 2 213x y z + + 5.CMR: a. 2 5 7 x x + + , xRb. 1 2 3 6 x y x y + + + + + , x, yRIII.CMR1.44a b c dabcd+ + + . (a, b , c, d 0) 2.33a b cabc+ + . (a, b , c 0) 3.1 1 1 9a b c a b c+ + + + (a, b , c > 0) -74-4.1 1 1 a b cbc ca ab a b c+ + + + (a, b , c > 0) 5.ab bc caa b cc a b+ + + +(a, b , c > 0) 6.2 21 12( ) x y x yx y+ + + +(x , y > 0)7. (a + b)(b+c)(c+a) 8abc (a, b , c 0) 8.1 1 1 8a b cb c a _ _ _+ + + , , , (a, b , c > 0) 9. (a + 2)(b + 8) (a + b) 32ab (a, b0) 10. (1 a)(1 b)(1 c) 8abc vi a + b + c = 1 v a, b, c 011.1 11 1 9x y _ _+ + , , vi x+y =1 v x , y > 0.12.(a + 2) (b + 8) 36 vi ab = 4 v a, b >013. 1 1 a b b a ab + a, b 114. 4 1 4 1 4 1 5 a b c + + + + + -5 (miny = 4 khi x = -1) 2.y = 92xx+ vi x > 2(miny = 8 khi x = 5) 3.y = 229xx+ vi x 0(miny = 6 khi x =3 t ) 4.y = 421 xx+ vi x 0 (miny = 2 khi x = t 1)5.y = (4 )(1 ) x xx+ + vi x > 0(miny = 9 khi x = 2)6.y = 2 4 x x + (miny = 2 khi 2 < x < 4)VII. Tm GTLN v GTNN ca biu thc S = xy + yz + zx bit x2 + y2 + z2 = 1-75-BI TP BT NG THCDng nh ngha:Chng minh cc bt ng thc sau1/ Cho a,b,c,d> 0a) nu a < b th b th>c) 1 < < 2 d)2 < < 32/ Cho 0, Chng minh rng ad) (a3 1)(a 1) 0e) 2abc a2 + b2c2f) (a + b)2 4ab g)a2 + ab + b2 0 h)a4 + b4 a3b + ab3i) 4ab(a b)2 (a2 b2)2 j) a2 + 2b2 + 2ab + b + 1 > 0k) l)2 + a2(1 + b2) 2a(1 + b)m) n) ( )2 o) ( )2p)+ b2 + c2 ab ac + 2bcq) a4 + b4 + c2 + 1 2a(ab2 a + c + 1) r) a4 + b4 + c2 + 1 2a(ab2 a + c + 1) s) 2a2 + 4b2 + c2 4ab + 2act)a2 + ab + b2 (a + b)2u) a + b + 2a2 + 2b2 2ab + 2b + 2av) (a + b + c)2 3(a2 + b2 + c2)4/ Cho a ,b [ 1;1] . Chng minh rng : |a + b| |1 + ab|a)Chng minh rng: nu x y 0th b)Chng minh rng: vi hai s a v b ty ta c + 5/ Cho a 2 , b 2. Chng minh rng : ab a + b6/ Cho x 0,chng minh rng: x4 + x + 1 > 07/ Cho ba s a ,b ,c [0;1],chng minh rng : a + b + c ab bc ca18/ Cho0 < a b c . Chng minh rng : b() + (a + c) ()(a + c)9/ Cho a > b > 0vc . Chng minh rng 10/ Choa + b + c 0. Chng minh rng : 011/ Cho ba s dng a ,b ,c ,chng minh rng :++12/ Cho cc s a,b,c,d tho a b c d 0. Chng minh rng :a)a2 b2 + c2 (a b + c)2b)a2 b2 + c2 d2 (a b + c d)2 13/ a) Cho a.b 1,Chng minh rng : b) Cho a 1, b 1 .Chng minh rng : c) Cho hai s x ,y thox + y 0.Chng minh rng : 14/ a,b,c,d chng minh rng a) b) 1 0 c) 1 d) a3 + b3 ab(a + b)e) a4 + a3b + ab + b2 4a2b f) (a + b)(1 + ab) 4ab g) (1 + a)(1 + b) (1 + )2h) i) j) +++j) (1 + a)(1 + b) (1 + )2h) 2 k) 3a2b3 16 l) 4m)2/ Cho a > 0 , chng minh rng :(1 + a)2 163/ Cho 3 s a ,b ,c > 0 ty . Chng minh rng: a) a2b + 2a b) a + b + c ( a2b + b2c + c2a +++ )4/ Cho 0 < a < b , chng minh rng: a bv ab = 1 ,chng minh rng : 214/*. Chng minh rng 15/ a) Chng minh rng nu b > 0 , c > 0 th : b)S dng kt qu trn chng minh rng nu a ,b ,c l ba s khng m c tnga + b + c = 1thb + c 16abc16/ Cho a + b = 1,Chng minh rng: ()() 917/ Cho a,b,c > 0 v a + b + c = 1 . Chng minh rng :a) ()()( ) 64 b) (a + b)(b + c)(c + a)abc 18*.Cho 4 s a ,b ,c ,d > 0 tho mn + + + 3Chng minh rngabcd 19/ Cho a,b,c l di ba cnh ca mt tam gic ,chng minh rng :a) ab + bc + ca < a2 + b2 + c2 0 , chng minh rng : a + 3.Khi no xy ra du = 26/ Cho hai sa 0 ; b 0 . Chng minh rng :a) 2 + 3 5 b) 17 12 5ab 17 b 12 a 5 +-78-c) 3a2b3 1627/ Chng minh rng 1.3.5.(2n 1) 331/*. n N chng minh rng :a) 1. . < 2) 1 n ( n1 n2+,_+b) 1.22.33.44nn < 2) 1 n ( n31 n 2+,_ +32/*.Cho m,n N ;m > n . Chng minh rng :( 1 + )m > ( 1 + )n33/*.Cho x1,x2,xn > 0 vx1 + x2 + .+ xn = 1Chng minh rng ()()( ) (n + 1)n34/*.Cho cc s x1, x2 ,y1, y2, z1, z2 tho mn x1.x2 > 0 ; x1.z1 y12 ;x2.z2 y22 Chng minh rng : (x1 + x2)(z1 + z2) (y1 + y2)235/*.Cho 3 s a ,b ,c (0;1). Chng minh rng trong 3 bt ng thc sau phi c mt bt ng thc sai:a(1 b) > 1/4(1); b(1 c) > 1/4 (2) ; c(1 a) > 1/4(3)36/*.Cho 3 s a,b,c > 0. Chng minh rng : ++ 37/** Cho x ,y ,z [0;1] ,chng minh rng : (2x + 2y + 2z)(2 x + 2 y + 2 z) (HBK 78trang 181,BT Trn c Huyn)38/*.Cho a , b , c > 1. Chng minh rng :a) 2b) 2 39/ Cho a ,b ,c >0,chng minh rng :a) b)c) 6d) ab + bc + cae) (a + b + c)(a2 + b2 + c2) 9abcf) a + b + cg) 40/ Cho ba s a ,b ,c tu . Chng minh rng : a2(1 + b2) + b2(1 + c2) + c2(1 +ab2) 6abc41/ Cho a ,b ,c > 0 tho : . Chng minh rng : 442/ Cho 3 s a, b, c tho a + b + c 1. Chng minh rng :a) ++ 9b)++ 9 43/ Cho a ,b ,c > 0 tho a + b + c k. Chng minh rng : ) 344/ Cho ba s a ,b ,c 0. Chng minh rng : 45/ Cho tam gic ABC,Chng minh rng : a) ha + hb + hc 9r b) < Dng tam thc bc hai1/ x , y R Chng minh rng :a) x2 + 5y2 4xy + 2x 6y + 3 > 0a) x2 + 4y2 + 3z2 + 14 >2x + 12y + 6z b) 5x2 + 3y2 + 4xy 2x + 8y + 9 0c) 3y2 + x2 + 2xy + 2x + 6y + 3 0-79-d) x2y4 + 2(x2 + 2)y2 + 4xy + x2 4xy3 e) (x + y)2 xy + 1 (x + y)f) 3 + 10 0g) (xy + yz + zx)2 3xyz(x + y + z)2/ Cho 4 s a ,b ,c ,d tho b< c < d chng minh rng : (a + b + c + d)2 > 8(ac + bd)3/Chng minh rng : (1 + 2x + 3x)2 < 3 + 3.4x + 32x+1 4/ Choax + by , x,y > 0. Chng minh rng :ab 1/45*/ Cho 1 x v < y 06**/ Cho a3 > 36v abc = 1.Xt tam thc f(x) = x2 ax 3bc +a) Chng minh rng :f(x) > 0 xb) Chng minh rng: + b2 + c2 > ab + bc + ca7/ Cho hai s x , y tho mn:x y . Chng minh rngx3 3x y3 3y + 4.Tm Gi tr nh nht ca cc hm s :a) y = x2 + b) y = x + 2 + vi x > 2c) y = x +vi x > 1d) y = vi x > 2e) y = vi x > 0f) y =+vi x (0;1)8/ Tm gi tr ln nht ca cc hm s sau:y = x(2 x)0 x2y = (2x 3)(5 2x) xy = (3x 2)(1 x) x 1y = (2x 1)(4 3x) x y = 4x3 x4 vix [0;4]11/ Trong mt phng ta Oxy,trn cc tia Ox v Oy ln lt ly cc im A v B thay i sao cho ng thng AB lun lun tip xc vi ng trn tm O bn knh R = 1. Xc nh ta ca A v B on AB c di nh nht 12/*.Cho a 3 ; b 4 ; c 2 .Tm gi tr ln nht ca biu thc A = 13/* Tm gi tr ln nht v gi tr nh nht ca hm s y = + -80-2 Bt phng trnh bc nhtI. Khi nim bt phng trnh mt n1. nh nghaCho hai hm sf(x),g(x) ccc tp xc nh Df,Dg. t Df Dg=D, mnh cha bin x D dng f(x)>g(x) gi l bt phng trnh mt n. V d: 2x+3>3x+6; 2x2+3x < 2x+5; 3x3+6x5x+3 2. Tp hp nghimTphpnghimcabt phngtrnhf(x) >g(x) ltphptt cccgitr x0) ( ) ( :0 0x g x f D > 3. iu kin ca bt phng trnhL iu kin ca n x sao cho f(x) v g(x) c nghaV d: iu kin ca bt phng trnh 23 1 x x x + + l 3x 0 v x+1 04. Bt phng trnh cha tham sL bt phng trnh cha cc ch ci khc ngoi n.V d: mx+2>5 (tham s m)5. H bt phng trnh mt nL h gm t hai bt phng trnh bc nht mt n. giimt hbtphngtrnhtagiitng btphngtrnhrily giaocc tp nghim .V d: Gii h 3 01 0xx '+ III. Bt phng trnh tng ng 1. nh ngha:hai bt phng trnh c gi l tng ngnhau nu chng c cng tp nghim. 2. nh l 2.1 nh l 1 (php cng, tr): Cho f(x) > g(x) xcnh trn D. Nu h(x) xc nh trn D th: f(x) > g(x) f(x) + h(x) > g(x) + h(x) *H qu: Nu chuyn mt biu thc t v ny sang v kia ca phng trnh v i du th ta c mt bt phng trnh mi tng ng vi phng trnh cho.2.2 nh l 2 (php nhn, chia): Cho f(x) > g(x) xc nh trn D + Nu h(x) xc nh trn D v h(x)>0 vi mi x D th bt phng trnh: f(x) > g(x) f(x).h(x) > g(x).h(x) + Nu h(x) xc nh trn D v h(x) g(x)f(x).h(x) < g(x).h(x)2.3. nh l 3 (bnh phng): Nu f(x) 0, g(x)0 thf(x) > g(x) f2(x) > g2(x)* Ch :Khi gii bt phng trnhcn lu cc vn sau+ t iu kin (nu c) trc khi bin i bt phng trnh.+ Khi nhn (chia) hai v bt phng trnh vi mt biu thc th ch xem biu thc m hay dng, hoc biu thc mang c hai gi tr m v dng.+ Khi qui ng mu s ca bt phng trnh: nu bit chc chn mu dng th khng i du.+ Nu f(x) 4 => tp nghim l T=(4;) +b) 2x-10 3x-2 -x8 x 8 => T=(] 8 ; * V d 2: Gii cc bt phng trnh saua) 2( 2)(2 1) 2 ( 1)( 3) x x x x x + + + p n: x1-81-b) 2 22 212 1x x x xx x+ + +>+ +p n: x +p n: x> * V d 2: Gii cc bt phng trnh saua) 5 2 3 4 3 314 4 6x x x x + > p n: 1/32000B ABBAB A ; '' 2000B ABBAB A B A B A < 0 T bt phng trnh ax+b > 0 ax > -b (1) Bin lun:+ Nu a = 0 => (1) 0x > -b . nu b > 0 => bptVSN . nu b < 0 => bptVN . nu b = 0 => bptVN+ Nu a > 0 => bpt c nghim x > ab+ Nu a < 0 => bpt c nghim x < abV d: gii v bin lun bt phng trnh(m-1)x -2+3m > 0 (1)Gii (1) (m-1)x > 2-3m (2)-82- . Num-1= 0 m=1(2) 0x > -1 => bpt VSN . Nu m-1> 0 m > 1 => bpt c nghimx > 13 2mm . Nu m-1 < 0 m < 1=> bpt c nghim x < 13 2mmKt lun:. m =1 bpt VN . m > 1 bpt c nghim x > 13 2mm . m < 1 bpt c nghim x < 13 2mm-83-BI TP1/ Giicc bt phng trnh saua)(2 3)( 1) x x x x + > + b)( 1 3)(2 1 5) 1 3 x x x + > c) 2( 4) ( 1) 0 x x + > d) 2( 2) ( 3) 0 x x + >e) 2(x1)+x > 333x ++f) 2 2( 2) ( 2) 2 x x + +g) x(7x)+6(x1)p s: a) S= [0;3) b) S= (;5) c) S=(1;4) (4;+) d) S= (3;+)e) S=(9/4;+); f) S=(; 2 / 4); g) (;6/11); h) S=[5;+); k) S=[3;2]l) S=(;4) (3;2) m) S={1}[2;;+) n) S=[21/4;13/2)2/ Gii cc h bt phng trnh sau:a) 3 5 2 14 1 3 2x xx x > '+ < +b)4 7 82 3 12x xx x '+ c) 5 2 4 55 4 2x xx x > + ' < +d) 2 1 3 45 3 8 9x xx x+ > + '+ e) 8 3 158 5 6 72 4 5 3x xx xx x < '+ < +f)1 222 3 64 3 2 5x x xx x+ + > +'+ > g) 6 5 2 46 2 4 33 2x xx x < ++ + '' > < 3. V d3.1 V d 1: gii phng trnh | x-1| + | 2x-4 | = 3(1) Gii Ta xt du cc biu thc x-1;2x-4 x 12 +x-1 - 0+ +2x-4 - - 0+ nhn vo bng xt du ta c: * nu x ) 1 ; ( th (1) -(x-1)-(2x-4)=3 -3x = -2 x = 32(nhn) * nu x ) 2 ; 1 [ th (1) x-1-(2x-4) = 3 x = 0 ) 2 ; 1 [ (loi) * nu x) ; 2 [ + th (1) x-1+2x-4 = 3 3x=8 x =38(nhn) Vy S =)'38;32 3.2 V d 2: gii cc bt phng trnh sau: a) | x-2 | > x+1b) | 2x+1 | < xTm tt l thuyt-86-1. Gii v bin lun phng trnh bc nht dngax + b >0ax > -b (1)Bin lun:+ Nu a = 0 th (1) 0.x > -b - nu b > 0 th bt phng trnh c v s nghim. - nu b 0 th bt phng trnh v nghim. + Nua > 0 th bpt c nghimx >ab. + Nu a < 0 th bpt c nghimxab < . Kt lun 2. Xt du nh thc bc nhtf(x) = ax+b (a0)x --b/a +f(x) Tri du a0Cng du a * Ch : Xt biu thc dng tch hoc thng cc nh thc bc nht ( v d : (ax+b)(cx+d)(fx+k);) )( () )...( )( (m kx h gxf ex d cx b ax+ ++ + +) ta xt du tt c cc nh th bc nht trn cng mt bng xt du. * Cc bc xt du biu thc :B1 : a biu thc cho v dng ax+b hoc dng tch hoc thng cc nh thc bc nht. B2 : Tm nghim cc nh thc bc nht. B3 : Xt du tt c cc nh thc trn cng mt bng xt du. B4 : Tng hp => kt lun. 3. Gii bt phng trnh bc nht B1 : a bt phng trnh v dngf(x)>0 hoc f(x) tp nghim. 4. Gii h gm 2 bt phng trnh bc nhtdng '(2) pt Bat(1) pt Bat (I) B1 : Gii bt phng trnh (1) => Tp nghimS1. B2 : Gii bt phng trnh (2) => Tp nghimS2 . B3 :Tp nghim S ca h (I) lS = S1S2. BI TP 11/ Xt du cc biu thc sau:a) f(x)= (2x1)(x+3) b) f(x)= (3x3)(x+2)(x+3)c) f(x)= 4 33 1 2 x x+ d) f(x)= 4x212/ Gii cc bt phng trnh sau2222 5 1 1) b) 1 2 1 1 ( 1)1 2 3 3 1) d) 14 3 1ax x x xx xcx x x x < + ++ < + +e) ( 2 x+2)(x+1)(2x3)>0 f) 4 133 1xx + +p s: a) S=(1;2] [3;+) b) S=(;1/2) [2/11;1)c) S=(;1) d) [52 6 3 2; + + 5+2 6 3 2 + + ]e) S=(;1) ( 2 ;3/2) f) S=[4/5;1/3)8/ Gii v bin lun bt phng trnha) mx+4>2x+m2b) 2mx+1x+4m2d) x(m21) < m41 e) 2(m+1)x (m+1)2(x1)9/Gii cc bt phng trnh sau-88-3 2) ( 3 2)( 1)(4 5) 0 b)0(3 1)( 4)3 1 2 2) 2d) 2 1 3 1 2 1xa x x xx xx x xcx x x + + > < + + + + p s: a) S=(;1) ( 2 3 /3;5/4) b) S=(1/3;3/2) hop (4;+)c) S= [3;1/2) d) S=(;1/3)[0;1/2)[8;+)10/ Gii h bt phng trnh 2 1( 3)( 2 ) 0)b) 2 1 34 33| | 12x xax xxxx > ' '< +

+ > + ' '+ < + < p s: a) S={4;5;6;7;8;9;10;11} b) S={1}12/ Gii cc phng trnh v bt phng trnh saua) |x+1|+|x1|=4 b) | 2 1| 1( 1)( 2) 2xx x>+ c) |5+x|+|x3|=8d) |x25x+6|=x25x+6 e) |2x1|= x+2 f) |x+2|+|x1|=5g) |3x5|2x3l) |x+1| |x|x+2p s: a) S={2;2} b) S= (4;1)(2;5) c) S=[5;3]d) S= x2 hoc x>3 e) S={1/3;3}f) S={3;2} g) S=(1;7/3)h) S=(4;1)(1;0] k) S=(;5/3)l)S=(;1]13. Giai bat phng trnh (cha gia tr tuyet oi) :12 34/ ; 6 2 6 3 4 /; 1 2 4 5 / ; 4 7 5 2 / ; 0 2 1 /2222+ + < + > + < x xx xe x x x x dx x c x x b x x a14. Giai bat phng trnh (cha can thc) : 1 3 2 / 4 2 2 3 / ; 2 5 /; 2 3 13 1 / ; 5 24 / ; 2 18 /2 2 22+ < + > > < +x x x f x x x e x x dx x c x x b x x a15/* Gii v bin lun phng trnha) (2x 2 )(xm)>0 b) 302 1xx m +16/* Gii v bin lun h phng trnha) ( 5)( 7 2 ) 00x xx m > ' b)2 51 2 10x xx m< ' -89-BI TP 2Bi 1: Gii v bin lun cc bt phng trnh sau theo tham s m a) m(x-m) x-1 b) mx+6 > 2x+3m c) (m+1)x + m < 3x+4 Bi 2: Gii cc bt phng trnh sau: a) 124 3>xxb) 125 2 xx c) 1 2512 x xd) x x x+1d) | x+2 | x+1p s: a) S=[4/3;6]b) V nghimc) S=(;1/2)d) S=RBi 4: Gii v bin lun cc phng trnh sau theo tham s m: a) | 2x-1 | = x+m b) | x-1 | =x+m Bi 5: Tm m cc bt phng trnh sau v nghim:a)m2x+4m-3 < x+m2 b)m2x+1 m+(3m-2)xBi 6: Gii cc h bt phng trnh sau a)' > > 435 ) 3 2 ( 228 155 8x xx b) ' >++ 214 3) 4 ( 2312 2 15xxx x p s: a) S={4;5;;11} b) S= {1}Bi 8: Tm s nguyn ln nht tho mn h bt phng trnh: '>> 95 4121181 4323 518) 2 ( 341 3x x xx x x p s: S= {4}-90-BI TP 31/Gii v bin lun cc bt phng trnh sau a)(m +1)2x > 2mx + mb)(m2+m)x - m2 - 2m0 c)(m+1)x 2m(x+1)+2+x. d) m2x-1 > x+m e) 1 m1 mx1 - m1 mx+>+ m 1 t f)2)x (m1 m1 x1 m1 - xx + ++>++m -1. 2/ Gii bt phng trnha) 2x2 - 5x + 2 > 0b) (x-2)2(x-4) < 0 c) -4 + x2 0d) 25(x+10)(-x+1) 0 e) 16x2 + 40x + 25 < 0f) 0) 1 (10>+ x xg) 91 534322++ xxx xh)191 2181+ ++ x xk) 0) 2 3 )( 2 (25< + x xl) 1 2211> x xm) 01322321++ xxn) 12 321+ x xxo) 11212+>+ ++xx xx3/ Gii cc h bt phng trnh sau a) '+ < ++ +19 2 3 47 2 1 3x xx x b) ' ++01) 4 2 )( 2 (113 2xx xxx c) ' +++41201 212x x xxxd) ' < >+2 2) 23 ( ) 19 5 (2121x xx x e) '++++++>++5 22 32 35 24321xxxxxxxxf) '+ +> + + +02) 2 )( 2 3 )( 1 (0 ) 5 )( 3 )( 2 (2xx x xx x xg) '+ < +09) 2 3 )( 4 (01 2) 3 1 )( 4 (22xx xxx x xh) '>+++ +01 2) 1 ( 312 1121123 2xx xxxx xx i)'+ > +) 1 ( 212 30 ) 25 9 )( 1 5 (22xxx xx xp s:a)S = [6 ; 8) b) S =(-; -4](1;2]c)S = (-;-1] (-21;+) d)S = (-1;2) e)S = (- 4; -25) f) S = (-2;- 3 ) [-1;2) [23;+)g)S = (-; -3) (23 ;-41)(0;31) (21;3) [4;+ ) h)S = (-1 ; - 21) (0;1) i)S = (- ; -3 ] [0 ; 53)-91-4/ Gii cc h bt phng trnh sau a) '< + > +0 1 3 20 ) 3 ( ) 2 (22x xx xb) '< + 0 ) 2 )( 1 (0 ) 1 (2x xxc) '++++>+5 231 22211 22xxxxxxxxd) '>++ ++ +1 221 32315 22xxxxxxx xe) ' + + > +0) 2 ( ) 7 () 6 ( ) 2 ( ) 1 (21 32 32 32x xx x xxxx xf) '++>++429324xxxxg)'+ + < > (3)5(2)3 3(1)0y xy xy x Gii Ta v cc ng thng (d1): x-y= 0(d2): x-3y+3= 0 (d3): x+y-5= 0 -93-I x(d3)(d1)(d2) -355011 Min I l min nghim. V d 2:Gii h000xyx y>>+ >' GiiV ccng thng : (d1): x= 0(d2): y= 0 (d3): x+y= 0xy-11S-94-BI TPBi 1: Gii cc bt phng trnh bc nht hai n a) x+3 +2(2y+5) < 2(1-x) b) 4(x-1) + 5(y-3) > 2x-9 c) 2x-y 3 d) 3+2y >0e) 2x-1 1 h) -3x+y+2 0k) 2x-3y+5 0Bi 2: Gii cc h bt phng trnh hai n a) '> + < > 53 30y xy xy xb) ' + +0423) 1 ( 20 13 2xyxy x d) ' +68 239 3yx yy xy xe) {3 02 3 1 0yx y < + >Bi3:Gi S l tp hp cc im trong mt phng to Oxy c to tho mn h bt phng trnh:' + + 052 22 2xy xy xy x .Tm cc im caS lm cho biu thc F = y-x t gi tr nh nht. Bi 4:Gi S l tp hp cc im trong mt phng to Oxy c to tho mn h bt phng trnh:2 01 02 1 0x yx yx y+ + ' + .Tm cc im caS lm cho biu thcF =2x+3y t gi tr max, min. -95-5 DU TAM THC BC HAI I/ Tam thc bc hai 1. nh ngha: Tam thc bc hai l biu thc c dng f(x) = ax2+bx+c(a0). 2. nh l (v du tam thc bc hai) Cho tam thc bc hai f(x)= ax2+bx+c (a0) v= b2-4ac + Nu< 0 th f(x) cng du vi h s a vi mi x. + Nu = 0 th f(x) cng du vi h s a vi ab2 .+ Nu > 0 th f(x) c hai nghim phn bit x1,x2( gi s x1< x2) : x0Cung dauhesoa -x1 x2+Dau cuaf(x)Cung dauhesoaTraidauhesoa0 * Ch : ta c th thaybi' V d 1: xt du cc tam thc sau a) f(x) = 3x2-2x+1b) f(x) = -4x2+12x-9 c) f(x) = x2-4x-5Gii a) cho f(x) = 0 3x2-2x+1 = 0.tnh' = -2 < 0 vy f(x) > 0x. b) cho f(x) = 0 -4x2+12x-9 = 0.tnh' = 0 vy f(x) < 0 23 x. c) cho f(x)= 0 x2-4x-5 = 0. tnh' = 9 => x1=-1 ;x2 = 5x0+- -1 5 + f(x) + _0vy f(x) > 0 ) ; 5 ( ) 1 ; ( + x f(x) < 0) 5 ; 1 ( xf(x) = 0 khi x= -1 , x = 5 V d 2: Xt du cc biu thc sau a)A = (2x2+9x+7)(x2+x-6) b)B = 222 5 73 10x xx x + + Giia)t 2x2+9x+7 = 0 27121xx x2+x-6 = 0 3221xx+ -+- + A x2+9x+7 +0--++x0 - -27 -3 -1 2+000000x2+x-6 + + - - +II/ Bt phng trnh bc hai 1. nh ngha: Bt phng trnh bc hai l bt phng trnh c mt trong cc dng sau: -96-ax2+bx+c > 0 ; ax2+bx+c < 0 ; ax2+bx+c 0 ax2+bx+c 0( a0).2 .Cch gii: gii bt phng trnh bc hai ta xt du tam thc bc hai , kt hp vi chiu ca bt phng trnh ta s tm c nghim ca bt phng trnh. V d 1: Gii cc bt phng trnh sau a) 3x2+2x+5 > 0 S=R b) -2x2+3x+5> 0 S=(-1;5/2) c) -3x2+7x-4 < 0S=(-;1) (4/3;+) d) 4x2-3x+1 0b)B = 222 5 73 10x xx x + +< 0V d 3. Xc nh m phng trnhx2+2(m+2)x-2m-1=0 c nghimHD:' =m2+6m+50 m5 hoc m1* Ch : Bi ton tm m f(x)= ax2+bx+c khng i du (>0, < ' ' III/ H bt phng trnh bc hai(10NC)1. nh ngha : L h gm t hai bt phng trnh bc hai tr ln. 2. Cch gii:- Gii bt phng trnh (1) tm c S1 - Gii bt phng trnh (2) tm c S2---------------------------------------------- - Gii bt phng trnh (n) tm c Sn Khi tp nghim ca h l: S = S1 S2Sn V d 1. Gii cc h bt phng trnh sau a) '< +> + +0 60 7 9 222x xx xGii Gii bpt(1) c S1 = ) ; 1 ( )27; ( + ;Gii bpt(2) dc S2 = (-3;2) Vy nghim ca h l S = S1S2= (-1;2) b) '< + > + 0 18 110 4 5 222x xx xV d 2. Tm m th bpt phng trnh sau (2m+1)x2+3(m+1)x+m+1 < 0 (*) v nghim. Gii + vi a = 0 m= 21 (*) 3102123 < < + x x .vy m =21 khng tho + vi a0 m21 khi phng trnh cho v nghim ' + + +> +' >0 ) 1 )( 1 2 ( 4 ) 1 ( 90 1 2002m m mm a ' >Smm1 521 vy khng c gi tr no ca m phng trnh v nghim.-97-* Ch : Bi ton tm m f(x)= ax2+bx+c khng i du (>0, '< '* Xt du cc nghim phng trnh bc hai Gi s phng trnh bc hai c hai nghim x1,x2 th: x1< 0 < x2 P < 0 (hai nghim tri du) x1x2 < 0 '< >000SP ( hai cng m) 0 < x1x2 '> >000SP(hai cng dng)BI TP 11/ Xt du cc tam thc bc hai saua) 2x2 +5x+2 b) 4x2 3x1 c) 3x2 +5x+1d) 3x2 +x+52/ Gii cc bt phng trnh saua) x2 2x+3>0b) x2 +9>6x c) 6x2 x2 0d) 13x2 +3x+6+h) 1 121x xx x+ + >i) 1 2 31 3 2 x x x+ 0 ; B > 0 ; C ; D; E; F )-142-BI 3 : n gin biu thcA = cotgxsinx-x sintgx =cosxx sin - 1=x sincosxsinx -x sinx cosx sin2= cosxB = x cos + x sinx cos + x sin3 3= sinx.cosx - 1 =x cos + x sinx cos + sinx.cosx - x )(sin x cos + x (sin2 2C =x tg - xcotgxsin - x cos2 22 2 = x cos . x sin =x cosx sin-x sinx cosx sin - x cos2 222222 2D = cosx - 1 . x cos + 1 = | x sin | = x cos - 12E = )2- tg(x- )2 5- x ( tg - )2 3+ x ( tg + ) x +2( tg= cotgx + x)] -2 tg[-( - ) 2 -2- tg(x - ) + x +2( tg= cotgx + tg cotgx + )2- tg(x - ) x +2( = cotgx cotgx + cotgx +cotgx = 0F =x) + x).tg(90 - cotg(90 - x) - (90 sin - 1) x + (90 cos - 10 00 20 2 = x cos - 1x sin - 122 tgx(cotgx) =x cos - 1x sin - 122+1 = x sin1= 1 +x sinx cos2 22 G = cos100 + cos300+...+cos1500 + cos1700 = (cos10 + cos170)+(cos30 + cos150)+(cos50 + cos130)+(cos70+cos110) + cos90 = (cos10cos10)+(cos30cos30)+(cos50cos50)+(cos70cos70) = 0H = sin2100 + sin2200 +...+ sin2900 = ( sin210 + sin280)+(sin220+sin270)+(sin230+sin260)+(sin240+sin250)+sin290= (sin210+cos210)+(sin220+cos220)+(sin230+cos230)+(sin240+cos240)+1 = 5K =sin2a - sin4acos4a - a 2 cos=a 3 tg =a sin . a 3 cos-a) sin3a.sin( -=22a - 4asin .22a + 4a2.cos24a - a 2sin .24a + 2a2.sin -L = sin2a.cotga cos2a= sin2a. 1 =a sina) - a 2 sin(=a sincos2a.sina - sin2a.cosa= cos2a -a sina cosM = tga + tg(a+3) +) + a ( tg3 2 -143- = tga + tga 3 + 13 - tga+.tga 3 - 13 + tga+ tga =3 2 tga.tg - 13 2tg + tga+3 tga.tg - 13tg + tga= a 3tg - 18tga+ tga2 =a 3 tg 3 =3tg2a - 1a) tg - tga 3 ( 33N = a cos +cosa - 1a sin .2acosa).tg + (122 2 = 1 = a cos + a sin = a cos +2atga sin2atg= a cos +2acos 22asin 2a sin2atg= a cos +cosa + 1cosa - 1a sin2atg2 2 222 22222 222 2P = tga - ga cottga + cotga = a tg - 1a tg + 1= tga -tga1tga +tga122= a 2 cos1=a cosa sin - a cosa cos122 22Q = (1 + 2cos2a + 2cos4a + 2cos6a).sina = sina + 2sina.cos2a + 2sina.cos4a + 2sinacos6a = sina +sin(a) + sin3a + sin(3a) + sin5a + sin(5a) + sin7a = sin7aS = a 5 cos + a 3 cos + a cosa 5 sin + a 3 sin + a sin =a 3 tg =) a 2 cos + 1 ( a 3 cos) a 2 cos + 1 ( a 3 sin=a 3 cos + a 2 cos a 3 cos 2a 3 sin + a 2 cos a 3 sin 2=a 3 cos + a 5 cos + a cosa 3 sin + a 5 sin + a sinR = cos10x + 2cos24x + 6cos3x.coxcosx8cosx.cos33x = cos10x + (1 + cos8x) cosx 2cosx(4cos33x3cosx)= cos10x + cos8x + 1 cosx 2cosx.cos9x= 2cos9x.cosx+1cosx2cos9x.cosx = 1cosxBI 4 : Chng minh ng thc lung gic a) (tg+ cotg )2 (tgcotg )2 = 4 VT = tg2+ cotg2 +2.tg .cotg (tg2 +tg2 2tg .cotg ) = 4b) g cot + tg tg + 1= cotg cotg + 1. tg + 1 tg2 242222. VT = tg = ) 1 + g cot1.( tg + 1 tg22 22c) sin4cos4= 2sin21. VT = (sin2 )2(cos2 )2 = (sin2 +cos2 )(sin2 cos2 ) = sin2 cos2=2sin21d) sin6cos6= 13sin2cos2 . VT = (sin2 +cos2 )(sin4 sin2 .cos2 +cos4 ) = sin4 sin2 .cos2 +cos4 = (sin2 +cos2 )2sin2 .cos2 sin2 .cos2e) (1 + tgx)(1 + cotgx ).sinx.cosx = 1 + 2sinx.cosx . VT =x cos . x sin )x cosx sin+x sinx cos+ 2 ( = x cos . x sin ). gx cot . tgx + tgx + gx cot + 1 (= 2sinx.cosx + cos2x + sin2x = 1 + 2cosx.sinx-144-f)sinxcosx - 1=x cos + 1x sin. sin2x = 1 cos2xg) tga . a 3 tg =a 2a.tg tg - 1a tg - a 2 tg2 22 2 .VT = atga 2 tg + 1 tga - tg2a. tg2a.tga - 1tga + a 2 tg=tg2a.tga) + 1 tg2a.tga)( - (1 tga) - tg2a )( tga + a 2 tg ( = tg3a.tgah) sin(a+b+c) =sina.cosb.cosc + cosa.sinb.cosc + cosa.cosb.sinc sina.sinb.sinc . VT = sin[(a+b)+c] = sin(a+b).cosc + cos(a+b).sinc = (sina.cosb+cosa.sinb)cosc + ( cosa.cosb sina.sina)sinc = sina.cosb.cosc + cosa.sinb.cosc + cosa.cosb.sinc sina.sinb.sinci) 8cos4a4cos2acos4a = 3 . VT = 8(cos2a)2 4cos2a cos4a = 2(1 + cos2a)2 4cos2a cos4a= 2 + 4cos2a + 2cos22a 4cos2a cos4a = 2 + 1 + cos4a cos4a = 3j)tg2a -a 2 cos1=a sin + a cossina - a cos.VT = a 2 cos1=cos2asin2a - 1=a sin - a cosa 2.cosa.sin - 1=sina) - sina)(cosa + (cosasina) - (cosa2 22tg2ak) )4- a ( g cot =sin2a - 1a 2 sin + 12 . VT =) a -4( ctog ).4+ a ( tg =) a -4sin( ).4+ a cos( 2) a -4cos( ).4+ a sin( 2=sin2a -2sina 2 sin +2sin =)4- a ( g cot = )]4 - cotg(a - ).[4- a +2( tg2l) sina + sinb +sinc = 2csin .2bcos .2acos 4 , bit a + b = c . VT = 2ccos2csin 2 +2b - acos2csin 2 =2ccos2csin 2 +2b - acos2b + asin 2 = 22bcos2acos2csin 4 = )2b + acos +2b - a(cos2csinm) cotgx + tgx = x 2 sin2 . VT = x 2 sin2=x cos . x sin . 22=x cos . x sin1=x cosx sin+x sinx cosn) cotgx cotg2x =x 2 sin1 . VT = x 2 sin1=x 2 sin . x sin) x - x 2 sin(=x 2 sin . x sincos2x.sinx - x cos . x 2 sin=sin2xcos2x-x sinx coso) 3 4cos2x + cos4x = 8sin4x . VP = 8sin4x = 2(1cos2x)2 = 24cos2x + 2cos22x = 24cos2x + 1 + cos4x= 3 4cos2x + cos4xp) sin4x + cos4x = 43+ x 4 cos41 . VT = (sin2x)2 + (cos2x)2 = 2 2)2x 2 cos + 1( + )2cos2x - 1(-145-=x 4 cos41+43=2x 4 cos + 1.21+21= x 2 cos21+212q) sin6x + cos6x = 85+ x 4 cos83 . VT = (sin2x+cos2x)3 3sin2xcos2x(sin2x+cos2x) = 1 3sin2x.cos2x = 85+ x 4 cos83=2cos4x - 1.43- 1 = x 2 sin43- 12r) cos3x.sin3x + sin3x.cos3x =x 4 sin43. VT = sin2x.sinx.cos3x + cos2x.cosx.sin3x = (1cos2x)sinx.cos3x + (1sin2x).cosx.sin3x = sinx.cos3x cos2xsinx.cos3x + cosx.sin3x sin2x.cosx.sin3x = sinx.cos3x + cosx.sin3x sinx.cosx(cosx.cos3x + sinx.sin3x) = sin(x+3x) sinx.cosx.cos(x3x) = sin4x 21sin2x.cos2x = sin4x 41sin4x = 43sin4xs) Sin5x 2sinx(cos4x + cos2x ) = sinx .VT = sin5x 2sinx.cos4x 2sinx.cos2x = sin5x [sin(3x) +sin5x][sin(x)+sin3x] = sin5x + sin3x sin5x + sinx sin3x = sinxt) x cos . x 2 cos =2xsin2x 7sin +2x 3cos2x 5cos . VT =)]2x+2x 7cos( - )2x-2x 7[cos(21+ )]23x+2x 5cos( + )23x-2x 5[cos(21=x cos . x 2 cos = ) x 3 cos + x (cos21= ) x 4 cos - x 3 (cos21+ ) x 4 cos + x (cos21u) x 3 sin41= ) x +3 x).sin( -3sin( . x sin . Ap dng tnh A = 18 13sin .18 7sin .18sin.VT=)21+ x2sin - 1 ( x sin21= ]3 2cos - x 2 .[cos21sinx. = x) -3sin( ) x +3sin( . x sin2 =x).sinx 4sin - 3 (412 = 41(3sinx4sin3x)= 41sin3x Ap dung : 81=6sin41=183 sin41=18 13sin .18 7sin .18sinv) sin(a+b)sin(ab) = cos2b cos2a . w) cos(a+b)cos(ab) =cos2a + cos2b 1 . x) sina + sinb + sinc sin(a+b+c) = 2a + csin2c + bsin2b + asin 4. y) cosa + cosb + cosc + cos(a+b+c) =2a + ccos2c + bcos2b + acos 4 . BI 5 : Chng biu thc lng gic c lp vi cc bin ( Khng ph thuc vo bin) A = gx cotsinx.cosx+x g cotx cos - xcotg22 2;B = .cos sin1 - g cot) cotg - (12 2 22 2 ; -146-C = x sin 4 + x cos + x cos 4 + x sin2 4 2 4 ; D = 3(sin8cos8 ) + 4(cos62sin6) + 6sin4;E = .cotg cotg - .sin sin sin - cos2 22 22 2;F = ) - (90 ).sin - (180 sin1 -) 90 - ( g cot)] + (90 cotg - 1 [0 2 0 2 0 22 0 2 ;( A=1; B =4; C= 3; D= 1; E=1; F =4 )G =) + x cos( ).3 - x cos( + x sin3 2 ; H =)3+ (x cos + xcos + )3- (xcos2 2 2 ; K =)3 2+ x ( sin + x sin + )3 2- x( sin2 2 2 ; (G= 1/4; H= 3/2 ; K=3/2)BI 6 : Tnh gi tr ca hm s lng gic Biu thc lng gic a) Bit cos= 4/5 v 00 < < 900 .+ Tnh sin, tg, cotg. + Tnh gi tr biu thc A = tg - g cot tg + g cot . ( sin= 3/5 ; A = 25/7 )b) Bit tg= 2 , vi l gc ca mt tam gic .+ Tnh cos , sin. + Tnh gi tr biu thc B = 2cos - sin cos 2 + sin . ( cos= 5 1 - ; B = 0)c) Cho tg+ cotg= 2 . ( 00 < < 900 )+ Tnh sin , cos, tg , cotg. + Tnh gi tr biu thcC = cotg + tg cos . sin2 2 .(= 450 ; C = 1/4 )d) Cho sin + cos= 2 . + Tnh sin , cos , g , cotg. + Tnh gi tr ca biu thc D = sin5+ cos5. ( cos= sin= 2 2 ; D = 4 / 2)e) Cho 3sin4cos4= 1/2 + Tnhbiu thc E = sin4+ 3cos4. ( E = 1 ) f) Bit tg750 =3 + 2 , tnh sin150, cos150 ; sin1050 , cos1050 .( cotg150 = 3 + 2; tg1050 = 3 - 2 -)g) Tnh 12cos ; 12n si; 12g cot ;12tg. ( 42 + 6= cos ;42 - 6= sin ) h) Cho cosa = 9/41 , vi < a < 3 /2. TnhF = tg( a /4) . ( F = 31/49 )i) Cho tgx = 1/2, tnh gi tr biu thc G = sin2x + tg2xsin2x - tg2x .( G = 1/4 )-147-j) Cho sina + cosa = 2 7 , tnhH = cos4a , I = tg2a. ( H = 1/8 ; I = 2 + 71 2)k) Cho cotgx = 3/4 , tnh gi tr biu thcJ = cos2x - x 2 sin + 1x 2 cos + x 2 sin + 1 . ( A = 3/4 )Bi 7 : Bin i biu thc lng gic v dng a) Bin i v dng tng A = sina.sin2a.sin3a = 21[cos(a2a)cos(a+2a)]sin3a = 21(cosa cos3a)sin3a = 21(cosa.sin3a cos3a.sin3a) = 21[21(sin4asin(2a))21sin6a] = 21[21(sin2a+sin4a)21sin6a]=41(sin2a+sin4asin6a) B = 4cosa.cos2a.sin2a 5 = 2(cosa+cos3a).sin2a 5 = 2cos3a.sin2a 5 + 2cosa.sin2a 5 = 2a 3sin +2a 7sin +2asin -2a 11sinb) Bini tng thnh tch C = cos3a sina = cos3a cos(2a) = )4+ a sin( ).4- a 2 sin( 2 D = 12cosa + cos2a = 2cosa + 2cos2a = 2(1cosa)cosa = 2.2.sin22a.cosa = 4 sin22a.cosa E = 1 + cosa+cos2a + cos3a = 2cos2a + (cos3a+cosa) = 2cos2a + 2cos2a.cosa = 2(cos2a+cos3a).cosa = a cos2acos2a 5cos 4-148-BI TP* Dung bang gia tr cac gia tr lng giac ac biet, va he thc c ban :( ) ( )( )( )( ) ( ) ( )x xxx xDxx xxx xCx x x B x x x x Ax xxx xly xy xy x ka a a a a jx xxxxixxxxh xxxgx xxxx xfxxxxex x x x d x x x x cx x x x b x x x x af ed cb ab a aab b aeb ab ab ad cb acos . cotsintan . coscotcos . sincotcos cotcot 1 cot sin 1 cot tan cot tancot . sin 1tantan sin)cot cottan tantan . tan )) tan 1 )( cos 1 ( tan cos sin 1 )sin2sincos 1cos 1sin)cos1tansin 1cos) tan 2 1sin 1sin 1)1 cos sincos 2cos 11 cos sin)cos 1sinsincos 1)sin . tan sin tan ) cot . cos cos cot )cos . sin 3 1 cos sin ) cos . sin 2 1 cos sin ))20 (31tan ) )23(178sin ))2(54cos ) )20 (32cot ))23(135cos ) ) 90 0 (54sin )4cos 26sin 22cos 50 sin 24cot3cos 2)45 tan 0 cos 2 30 sin 245 tan 90 sin)3cos 83cot 26sin 3 )6tan 33sin 2 cos )2cot 7 tan 22cos 3 0 sin 5 )2 22 22 2 2 2 22222 2 2 2 2 2 2 22 2 6 6 2 2 4 40 02323 320 0 0 220202222 2 2 + + + + ++++ + + + + +++ +++ ++ ++ + +<