tuyển chọn 410 hệ phương trình và các phương pháp giải hệ phương trình

5/25/2018 Tuy n ch n 410 h ph ng tr nh v c c ph ng ph p gi i h ph ng tr nh

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n

Nguyn Minh TunSinh vin K62CLC - Khoa Ton Tin HSPHN

TUYN CHN 410 BI HPHNG TRNH I S

BI DNG HC SINH GII V LUYN THI I HC - CAONG


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n

H Ni, ngy 9 thng 10 nm 2013


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nMc lc

Li ni u 4

1 Mt s phng php v cc loi h c bn 5

1.1 Cc phng php chnh gii h phng trnh . . . . . . . . . . . . . . . . . . 51.2 Mt s loi h c bn. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6

2 Tuyn tp nhng bi h c sc 7

2.1 Cu 1 n cu 30 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

2.2 Cu 31 n cu 60 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23

2.3 Cu 61 n cu 90 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38

2.4 Cu 91 n cu 120 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50

2.5 Cu 121 n cu 150 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 65

2.6 Cu 151 n cu 180 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 82

2.7 Cu 181 n cu 210 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 99

2.8 Cu 211 n cu 240 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 114

2.9 Cu 241 n cu 270 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 131

2.10 Cu 271 n cu 300 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 149

2.11 Cu 301 n cu 330 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 168

2.12 Cu 331 n cu 360 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 185

2.13 Cu 361 n cu 390 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 201

2.14 Cu 391 n cu 410 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 218

Ti liu tham kho 228


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inhTu

nLi ni u

H phng trnh i s ni chung v h phng trnh i s hai n ni ring l mt phnquan trng ca phn i s ging dy THPT . N thng hay xut hin trong cc k thi hcsinh gii v k thi tuyn sinh i hc - Cao ng.

Tt nhin gii tt h phng trnh hai n khng phi n gin . Cn phi vn dng ttcc phng php, hnh thnh cc k nng trong qu trnh lm bi. Trong cc k thi i hc, cu

h thng l cu ly im 8 hoc 9.

y l mt ti liu tuyn tp nhng kh dy nn ti trnh by n di dng mt cun schc mc lc r rng cho bn c d tra cu. Cun sch l tuyn tp khong 400 cu h c sc,t n gin, bnh thng, kh, thm ch n nh v kinh in. c bit, y hon ton lh i s 2 n. Ti mun khai thc tht su mt kha cnh ca i s. Nu coi Bt ng thc3 bin l phn p nht ca Bt ng thc, mang trong mnh s uy nghi ca mt ng hong thH phng trnh i s 2 n li mang trong mnh v p gin d, trong sng ca c gi thnqu lm say m bit bao g si tnh.

Xin cm n cc bn, anh, ch, thy c trn cc din n ton, trn facebook ng gp vcung cp rt nhiu bi h hay. Trong cun sch ngoi vic a ra cc bi h ti cn lng thmmt s phng php rt tt gii. Ngoi ra ti cn gii thiu cho cc bn nhng phng phpc sc ca cc tc gi khc . Mong y s l mt ngun cung cp tt nhng bi h hay chogio vin v hc sinh.

Trong qu trnh bin son cun sch tt nhin khng trnh khi sai st.Th nht, kh nhiubi ton ti khng th nu r ngun gc v tc gi ca n. Th hai : mt s li ny sinh trongqu trnh bin son, c th do li nh my, cch lm cha chun, hoc trnh by cha p dokin thc v LATEX cn hn ch. Tc gi xin bn c lng th. Mong rng cun sch s honchnh v thm phn s. Mi kin ng gp v sa i xin gi v theo a ch sau y :

Nguyn Minh TunSinh Vin Lp K62CLC

Khoa Ton Tin Trng HSP H NiFacebook :https://www.facebook.com/popeye.nguyen.5

S in thoi : 01687773876Nick k2pi, BoxMath : Popeye
http://%20https//www.facebook.com/popeye.nguyen.5http://%20https//www.facebook.com/popeye.nguyen.5http://%20https//www.facebook.com/popeye.nguyen.5http://%20https//www.facebook.com/popeye.nguyen.5http://%20https//www.facebook.com/popeye.nguyen.5http://%20https//www.facebook.com/popeye.nguyen.5http://%20https//www.facebook.com/popeye.nguyen.5


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nChng 1Mt s phng php v cc loi h cbn

1.1 Cc phng php chnh gii h phng trnhI. Rt x theo y hoc ngc li t mt phng trnh

II. Phng php th1. Th hng s t mt phng trnh vo phng trnh cn li2. Th mt biu thc t mt phng trnh vo phng trnh cn li3. S dng php th i vi c 2 phng trnh hoc th nhiu ln.

III. Phng php h s bt nh1. Cng tr 2 phng trnh cho nhau2. Nhn hng s vo cc phng trnh ri em cng tr cho nhau.3. Nhn cc biu thc ca bin vo cc phng trnh ri cng tr cho nhau

IV. Phng php t n ph

V. Phng php s dng tnh n iu ca hm s

VI. Phng php lng gic ha

VII. Phng php nhn chia cc phng trnh cho nhau

VIII. Phng php nh gi1. Bin i v tng cc i lng khng m2. nh gi s rng buc tri ngc ca n, ca biu thc, ca mt phng trnh3. nh gi da vo tam thc bc 24. S dng cc bt ng thc thng dng nh gi

IX. Phng php phc ha

X. Kt hp cc phng php trn


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6 Mt s phng php v cc loi h c bn

1.2 Mt s loi h c bnA. H phng trnh bc nht 2 n

I. Dng ax + by= c (a2 + b2 = 0)

ax + by=c (a2

+ b2

= 0)II. Cch gii1. Th2. Cng i s3. Dng th4. Phng php nh thc cp 2

B. H phng trnh gm mt phng trnh bc nht v mt phng trnh bc hai

I. Dng ax2 + by2 + cxy+ dx + ey+ f= 0

ax + by=c

II. Cch gii:Th t phng trnh bc nht vo phng trnh bc hai

C. H phng trnh i xng loi II. Du hiui vai tr ca xvycho nhau th h cho khng iII. Cch gii:Thng ta s t n ph tng tch x + y = S, xy=P (S2 4P)

D. H phng trnh i xng loi II

I. Du hiui vai tr ca xvycho nhau th phng trnh ny bin thnh phng trnh kiaII. Cch gii:Thng ta s tr hai phng trnh cho nhau

E. H ng cpI. Du hiu

ng cp bc 2

ax2 + bxy+ cy2 =d

ax2 + bxy+ cy2 =d

ng cp bc 3ax3 + bx2y+ cxy2 + dy3 =e

ax3 + bx2y+ cxy2 +dy3 =e

II. Cch gii:Thng ta s t x= ty hoc y = tx

Ngoi ra cn mt loi h na ti tm gi n l bn ng cp, tc l hon ton c th av dng ng cp c .Loi h ny khng kh lm, nhng nhn nhn ra c n cn phikho lo sp xp cc hng t ca phng trnh li. Ti ly mt v d n gin cho bn c

Gii h :

x3 y3 = 8x + 2yx2 3y2 = 6

Vi h ny ta ch vic nhn cho v vi v s to thnh ng cp. V khi ta c quynchn la gia chia c 2 v cho y3 hoc t x= ty

Nguyn Minh Tun - K62CLC Ton Tin - HSPHN. My facebook : Popeye Nguyn


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nChng 2Tuyn tp nhng bi h c sc

2.1 Cu 1 n cu 30

Cu 1

(x y) (x2 + y2) = 13(x + y) (x2 y2) = 25

Gii

D dng nhn thy y l mt h ng cp bc 3, bnh thng ta c nhn cho ln ri chia 2v cho x3 hoc y3. Nhng hy xem mt cch gii tinh t sau y:Ly(2) (1)ta c : 2xy(x y) = 12 (3)Ly(1) (3)ta c : (x y)3 = 1 x= y + 1V sao c th c hng ny ? Xin tha l da vo hnh thc i xng ca h. Ngon lnh

ri. Thay vo phng trnh u ta c

(y+ 1)2 + y2 = 13

y = 2y = 3

Vy h cho c nghim (x; y) = (3; 2), (2;3)

Cu 2 x3 8x= y3 + 2y

x

2

3 = 3 (y2

+ 1)

Gii

nh sau : Phng trnh 1 gm bc ba v bc nht. Phng trnh 2 gm bc 2 v bc 0(hng s).R rng y l mt h dng na ng cp. Ta s vit li n a v ng cpH cho tng ng :

x3 y3 = 8x + 2yx2

3y2 = 6

Gi ta nhn cho hai v a n v dng ng cp

6 x3 y3 = (8x + 2y) x2 3y2 2x (3y x) (4y+ x) = 0


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n

8 Tuyn tp nhng bi h c sc

TH1 : x= 0thay vo (2) v nghimTH2 : x= 3ythay vo (2) ta c:

6y2

= 6 y= 1, x= 3y= 1, x= 3TH3 : x= 4ythay vo (2) ta c:

13y2 = 6

y=

6

13, x= 4

6

13

y=

6

13, x= 4

6

13

Vy h cho c nghim :(x; y) = (3; 1), (3;1),4

6

13 ;6

13

,

46

13 ;6

13

Cu 3

x2 + y2 3x + 4y= 13x2 2y2 9x 8y = 3

Gii

khi nhn 3 vo PT(1) ri tr i PT(2) s ch cn y . Vy

3.P T(1) P T(2) y2 + 4y= 0

y= 0 x=

372

y= 4 x= 3

7

2

Vy h cho c nghim : (x; y) =

3 7

2 ; 0

,

37

2 ;4

Cu 4

x2 + xy+y2 = 19(x y)2x2 xy+y2 = 7 (x y)

Gii

Nhn xt v tri ang c dng bnh phng thiu, vy ta th thm bt a v dng bnhphng xem sao. Nn a v (x y)2 hay (x+ y)2. Hin nhin khi nhn sang v phi ta schn phng n u

H cho tng ng

(x y)2 + 3xy= 19(x y)2(x y)2 + xy= 7 (x y)

t x y = a vxy= b ta c h mi



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n

2.1 Cu 1 n cu 30 9

b= 6a2

a2 + b= 7a

a= 0, b= 0a= 1, b= 6

x y = 0xy= 0

x y = 1xy= 6

x= 0, y= 0x= 3, y= 2x= 2, y = 3

Vy h cho c nghim :(x; y) = (0; 0) , (3;2)(2;3)

Cu 5

x3 + x3y3 + y3 = 17x +xy+ y= 5

GiiH i xng loi I ri. No problem!!!

H cho tng ng (x + y)3 3xy(x + y) + (xy)

3 = 17(x + y) +xy = 5

t x + y= avxy= bta c h mi

a3 3ab + b3 = 17a +b= 5

a= 2, b= 3a= 3, b= 2

x + y= 2xy= 3

x + y= 3xy= 2

x= 2, y= 1x= 1, y= 2

Vy h cho c nghim (x; y) = (1; 2), (2; 1)

Cu 6

x(x + 2)(2x + y) = 9x2 + 4x + y = 6

Giiy l loi h t n tng tch rt quen thuc

H cho tng ng

(x2 + 2x) (2x +y) = 9(x2 + 2x) + (2x + y) = 6

t x2 + 2x= av2x + y = b ta c h mi ab= 9a + b= 6

a= b = 3

x2 + 2x= 32x + y = 3

x= 1, y = 1x= 3, y= 9

Vy h cho c nghim (x; y) = (1; 1), (3;9)

Cu 7

x +y xy= 3x + 1 +

y+ 1 = 4

GiiKhng lm n g c c 2 phng trnh, trc gic u tin ca ta l bnh phng ph skh chu ca cn thc



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n


(2) x + y+ 2 + 2

xy+ x + y+ 1 = 16

M t (1) ta c x + y= 3 + xynn

(2) 3 + xy+ 2 + 2xy+ xy+ 4 = 16 xy= 3 xy= 9x +y = 6 x= y = 3Vy h cho c nghim (x; y) = (3; 3)

Cu 8

x + 5 +

y 2 = 7x 2 + y+ 5 = 7

Giii xng loi II. Khng cn g ni. Cho 2 phng trnh bng nhau ri bnh phng tungte ph s kh chu ca cn thciu kin : x, y 2T 2 phng trnh ta c

x + 5 +

y 2 = x 2 +

y 5

x + y+ 3 + 2

(x + 5)(y 2) =x + y+ 3 + 2

(x 2)(y+ 5)

(x + 5)(y

2) = (x

2)(y+ 5)

x= y

Thay li ta c

x + 5 +

x 2 = 7 x= 11

Vy h cho c nghim : (x; y) = (11;11)

Cu 9 x2 + y2 +

2xy= 8

2

x +y = 4

Gii

H cho c v l na i xng na ng cp, bc ca PT(2) ang nh hn PT(1) mtcht. Ch cn php bin i bnh phng (2) s va bin h tr thnh ng cp va ph bbt i cniu kin : x, y 0H cho

2(x2 + y2) + 2xy= 16x +y+ 2xy= 16 2 (x2 + y2) =x + y x= yThay li ta c : 2

x= 4 x= 4



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2.1 Cu 1 n cu 30 11

Vy h cho c nghim (x; y) = (4; 4)

Cu 10

6x

2

3xy+x= 1 yx2 + y2 = 1

Gii

Mt cch trc gic khi nhn thy h cha tam thc bc 2 l th xem liu c phn tch cthnh nhn t hay khng ? Ta s th bng cch tnh theo mt n c chnh phng haykhng. Ngon lnh l PT(1) xp nh tin.Phng trnh u tng ng (3x 1)(2x y+ 1) = 0Vi x=

1

3 y= 2

2

3

Vi y= 2x + 1 x2 + (2x + 1)2 = 1

x= 0, y = 1

x= 45

, y=3

5

Vy h cho c nghim (x; y) =

1

3;2

2

3

, (0, 1),

4

5;3

5

Cu 11 x 2y xy= 0

x

1 +

4y

1 = 2

Gii

Phng trnh u l dng ng cp riiu kin x 1, y1

4T phng trnh u ta c :

x +

y

x 2y = 0 x= 4yThay vo (2) ta c

x 1 + x 1 = 2 x= 2

Vy h cho c nghim (x

;y

) =

2;

1

2

Cu 12

xy+ x +y = x2 2y2x

2y yx 1 = 2x 2y

Gii

iu kin : x

1, y

0

Phng trnh u tng ng

(x + y) (2y x + 1) = 0

x= yx= 2y+ 1



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n


Vi x= yloi v theo iu kin th x, yphi cng duVi x= 2y+ 1 th phng trnh 2 s tng ng

(2y+ 1)

2y y

2y = 2y+ 2

2y(y+ 1) = 2y+ 2 y= 2 x= 5


Cu 13

x + 1 +

y+ 2 = 6x +y = 17

Gii

iu kin x, y 1H cho tng ng

x + 1 + y+ 2 = 6(x + 1) + (y+ 2) = 20t

x + 1 =a 0,y+ 2 =b 0. H cho tng ng

a + b= 6a2 +b2 = 20

a= 4, b= 2a= 2, b= 4

x= 15, y = 2x= 3, y= 14

Vy h cho c nghim (x; y) = (15; 2), (3;14)

Cu 14

y2 = (5x + 4)(4 x)y2 5x2 4xy+ 16x 8y+ 16 = 0

Gii

Phng trnh 2 tng ng

y2 + (5x + 4)(4 x) 4xy 8y= 0 2y2 4xy 8y = 0

y= 0y= 2x + 4

Vi y= 0th suy ra : (5x + 4) (4 x) = 0 x= 4

x= 45

Vi y= 2x + 4th suy ra (2x + 4)2 = (5x + 4)(4 x) x= 0Vy h cho c nghim (x; y) = (4; 0),

4

5; 0

, (0; 4)

Cu 15

x2 2xy+x +y = 0x4 4x2y+ 3x2 + y2 = 0

Gii

H cho tng ng



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n

2.1 Cu 1 n cu 30 13

x2 + y=x(2y 1)(x2 + y)

2+ 3x2 (1 2y) = 0 x

2(2y 1)2 + 3x2(2y 1) = 0 x2(2y 1)(2y 4) = 0

x= 0, y = 0

y=1

2(L)

y= 2, x= 1 2Vy h cho c nghim (x; y) = (0; 0), (1; 2), (2; 2)

Cu 16 x + y+ xy(2x +y) = 5xyx + y+ xy(3x

y) = 4xy

Gii

P T(1) P T(2) xy(2y x) =xy

xy= 0x= 2y 1

Vi xy= 0 x + y= 0 x= y = 0Vi x= 2y 1

(2y 1) +y+ (2y 1)y(5y 2) = 5(2y 1)y y= 1, x= 1

y=

9

41

20 , x= 1 +

41

10

y=9 +

41

20 , x=

41 1

10

Vy h cho c nghim(x; y) = (0; 0), (1; 1),

1 +

41

10 ;

9 4120

,

41 1

10 ;

9 +

41

20

Cu 17 x2 xy+y2 = 32x3 9y3 = (x y)(2xy+ 3)

Gii

Nu ch xt tng phng trnh mt s khng lm n c g. Nhng 2 ngi ny b rngbuc vi nhau bi con s 3 b n. Php th chng ? ng vy, thay 3 xung di ta s ra mtphng trnh ng cp v kt qu p hn c mong iTh 3 t trn xung di ta c

2x3

9y3 = (x

y) x2 +xy+ y2 x3 = 8y3 x= 2y

(1) 3y2 = 3 y = 1, x= 2Vy h cho c nghim (x; y) = (2; 1), (2;1)



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Cu 18

x + y+

x y = 1 +

x2 y2x +

y = 1

Gii

iu kin :x y 0Phng trnh u tng ng

x +y 1 = x y x + y 1 x + y= 1

x y = 1

x=

1 yx=

1 +y

T

1 y+ y= 1y+ 1 +

y= 1

y= 0, x= 1y= 1, x= 0(L)

y= 0, x= 1


Cu 19

2x y = 1 +x(y+ 1)x3 y2 = 7

Gii

iu kin : x(y+ 1) 0T (2) d thy x >0 y 1(1)

x

y+ 1 2x +y+ 1 = 0 x= y + 1 (y+ 1)3 y2 = 7 y = 1, x= 2Vy h cho c nghim (x; y) = (2; 1)

T cu 20 tr i ti xin gii thiu cho cc bn mt phng php rt mnh gii quyt gn p rt nhiu cc h phng trnh hu t. gi h s bt nh(trong y ti s gi n bng tn khc : UCT). S mt khong hn chc v d din t trn vn phng php ny

Trc ht im qua mt mo phn tch nhn t ca a thc hai bin rt nhanh bng mytnh Casio. Bi vit ca tc gi nthoangcute.

V d 1 : A= x2 + xy 2y2 + 3x + 36y 130Thc ra y l tam thc bc 2 th c th tnh phn tch cng c. Nhng th phn tchbng Casio xem .Nhn thy bc ca x v y u bng 2 nn ta chn ci no cng cCho y = 1000ta c A= x2 + 1003x 1964130 = (x + 1990) (x 987)Cho 1990 = 2y 10 v 987 = y 13A= (x + 2y

10)(x

y+ 13)

V d 2 : B= 6x2y 13xy2 + 2y3 18x2 + 10xy 3y2 + 87x 14y+ 15



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n

2.1 Cu 1 n cu 30 15

Nhn thy bc ca x nh hn, cho ngay y = 1000B= 5982x2 12989913x + 1996986015 = 2991 (2x 333)(x 2005)Cho 2991 = 3y 9 ,333 =

y 13

, 2005 = 2y + 5

B= (3y 9)2x y

1

3

(x 2y 5) = (y 3)(6x y+ 1) (x 2y 5)

V d 3 : C=x3 3xy2 2y3 7x2 + 10xy+ 17y2 + 8x 40y+ 16Bc ca xvynh nhauCho y = 1000ta c C=x3 7x2 2989992x 1983039984Phn tch C= (x 1999)(x + 996)2Cho 1999 = 2y 1v996 =y 4C= (x 2y+ 1) (x + y 4)2

V d 4 : D= 2x2y2 + x3 + 2y3 + 4x2 +xy+ 6y2 + 3x + 4y+ 12Bc ca xvynh nhauCho y = 1000ta c D= (x + 2000004) (x2 + 1003)Cho 2000004 = 2y2 + 4v1003 =y + 3D= (x + 2y2 + 4) (x2 +y+ 3)

V d 5 : E=x3y+ 2x2y2 + 6x3 + 11x2y xy2 6x2 7xy y2 6x 5y+ 6Bc ca ynh hn

Cho x = 1000 ta c E = 1998999y2

+ 1010992995y + 5993994006 =2997 (667y+ 333333) (y+ 6)

o ha E=999 (2001y+ 999999) (y+ 6)Cho 999 =x 1, 2001 = 2y+ 1, 999999 =x2 1E= (x 1) (y+ 6) (x2 + 2xy+y 1)

V d 6 : F = 6x4y+ 12x3y2 + 5x3y 5x2y2 + 6xy3 +x3 + 7x2y+ 4xy2 3y3 2x2 8xy+3y2 2x + 3y 3Bc ca y nh hnCho x= 1000ta c F= 5997y3 + 11995004003y2 + 6005006992003y+ 997997997Phn tch F= (1999y+ 1001001) (3y2 + 5999000y+ 997)Cho 1999 = 2x 1, 1001001 =x2 + x + 1, 5999000 = 6x2 x, 997 =x 3F = (x2 + 2xy+ x y+ 1) (6x2y xy+ 3y2 +x 3)

Lm quen c ri ch ? Bt u no

Cu 20

x2 +y2 =1

5

4x2 + 3x5725

= y(3x + 1)

Gii



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n


Li gii gn p nht ca bi trn l

25.P T(1) + 50.P T(2) (15x + 5y 7)(15x + 5y+ 17) = 0

n y d dng tm c nghim ca h : (x; y) = 25

;1

5 ,11

25

; 2

25

Cu 21

14x2 21y2 6x + 45y 14 = 035x2 + 28y2 + 41x 122y+ 56 = 0

Gii

Li gii gn p nht ca bi ny l

49.P T(1) 15.P T(2) (161x 483y+ 218)(x + 3y 7) = 0V n y cng d dng tm ra nghim (x; y) = (2;3), (1; 2)

Qua 2 v d trn ta t ra cu hi : V sao li th ? Ci nhm thnh nhn t th ti khngni bi t hn cc bn c n trn ri. V sao y l ti sao li ngh ra nhng hng skia nhn vo cc phng trnh, mt s tnh c may mn hay l c mt phng php. Xin tha chnh l mt v d ca UCT. UCT l mt cng c rt mnh c th qut sch gn nh tonb nhng bi h dng l hai tam thc. Cch tm nhng hng s nh th no. Ti xin trnhby ngay sau y. Bi vit ca tc gi nthoangcute.

Tng Qut:

a1x2 +b1y

2 + c1xy+ d1x + e1y+ f1 = 0a2x

2 +b2y2 + c2xy+ d2x + e2y+ f2 = 0

Gii

Hin nhin nhn xt y l h gm hai tam thc bc hai. M nhc n tam thc th khngth khng nhc ti mt i tng l . Mt tam thc phn tch c nhn t hay khngphi xem x hoc y ca n c chnh phng hay khng. Nu h loi ny m t ngay mtphng trnh ra k diu th chng ni lm g, th nhng c hai phng trnh u ra rtk cc th ta s lm nh no. Khi UCT s ln ting. Ta s chn hng s thch hp nhn vomt (hoc c hai phng trnh) p sao cho chnh phng.

Nh vy phi tm hng s k sao cho P T(1) +k.P T(2)c th phn tch thnh nhn tt a= a1+ ka2, b= b1+ kb2, c= c1+kc2, d= d1+kd2, e= e1+ke2, f=f1+ kf2S kl nghim ca phng trnh sau vi a = 0

cde+ 4abf=ae2 +bd2 +f c2

D vng c hn mt cng thc gii h phng trnh loi ny. Tc gi ca n kh xutsc !!!. Th kim chng li v d 21 nha= 14 + 35k, b=

21 + 28k, c= 0, d=

6 + 41k, e= 45

122k, f=

14 + 56k

S ks l nghim ca phng trnh

4(14+35k)(21+28k)(14+56k) = (14+35k)(45122k)2+(21+28k)(6+41k)2 k= 1549



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2.1 Cu 1 n cu 30 17

Nh vy l P T(1) 1549

.P T(2)hay 49.P T(1) 15.P T(2)Mt cht lu l khng phi h no cng y cc hng s. Nu khuyt thiu phn no thcho hng s l 0. Ok!!Xong dng ny ri. Hy lm bi tp vn dng. y l nhng bi h ti tng hp t nhiu

ngun.

1.

x2 + 8y2 6xy+ x 3y 624 = 021x2 24y2 30xy 83x + 49y+ 585 = 0

2.

x2 + y2 3x + 4y= 13x2 2y2 9x 8y= 3

3.

y2 = (4x + 4)(4 x)y2 5x2 4xy+ 16x 8y+ 16 = 0

4. xy 3x 2y= 16x2

+ y2

2x 4y= 335.

x2 + xy+ y2 = 3x2 + 2xy 7x 5y+ 9 = 0

6.

(2x + 1)2 + y2 + y= 2x + 3xy+x= 1

7.

x2 + 2y2 = 2y 2xy+ 13x2 + 2xy y2 = 2x y+ 5

8.

(x 1)2 + 6(x 1)y+ 4y2 = 20x2 + (2y+ 1)2 = 2

9. 2x2 + 4xy+ 2y2 + 3x + 3y 2 = 0

x2 + y2 + 4xy+ 2y= 0

10.

2x2 + 3xy= 3y 133y2 + 2xy= 2x + 11

11.

4x2 + 3y(x 1) = 73y2 + 4x(y 1) = 3

12.

x2 + 2 =x(y 1)y2 7 =y(x 1)

13.

x2 + 2xy+ 2y2 + 3x= 0xy+ y2 + 3y+ 1 = 0

Cu 22

x3 y3 = 352x2 + 3y2 = 4x 9y

Gii

Li gii ngn gn cho bi ton trn l

P T(1) 3.P T(2) (x 2)3 = (y+ 3)3 x= y + 5

Thay vo (2) ta d dng tm ra nghim(x; y) = (2;3), (3;2)Cu hi t ra y l s dng UCT nh th no ? Tt nhin y khng phi dng trn nari. Trc ht nh gi ci h ny



NguynM

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- Bc ca xvyl nh nhau- Cc bin x,yc lp vi nhau- Phng trnh mt c bc cao hn PT(2)Nhng nhn xt trn a ta n tng nhn hng s vo PT(2) P T(1) +a.P T(2)ac v dng hng ng thc A3 =B3P T(1) +a.P T(2) x3 + 2ax2 4ax y3 + 3ay2 + 9ay 35 = 0Cn tm asao cho v tri c dng (x + )3 (y+ )3 = 0

Cn bng ta c :

3 3 = 353= 2a32 = 4a

a= 3= 2= 3

VyP T(1) 3.P T(2) (x 2)3 = (y+ 3)3OK ?? Th mt v d tng t nh

Gii h: x3 + y3 = 91

4x2

+ 3y2

= 16x

+ 9y

Gi : P T(1) 3.P T(2) (x 4)3 = (y+ 3)3

Cu 23

x3 + y2 = (x y)(xy 1)x3 x2 + y+ 1 =xy(x y+ 1)

Gii

Hy cng ti phn tch bi ton ny. Tip tc s dng UCTnh gi h :-Bc ca x cao hn bc ca y-Cc bin x,y khng c lp vi nhau-Hai phng trnh c bc cao nht ca x v y nh nhauV bc x ang cao hn bc y v bc ca y ti 2 phng trnh nh nhau nn ta hy nhn tungri vit li 2 phng trnh theo n y. C th nh sau :

y2 (x + 1) y (x2 + 1) +x3 + x= 0y2x y (x2 +x 1) +x3 x2 + 1 = 0

By gi ta mong c rng khi thay x bng 1 s no vo h ny th s thu c 2 phngtrnh tng ng. Tc l khi cc h s ca 2 phng trnh s t l vi nhau . Vy :

x + 1

x =

x2 + 1

x2 + x 1= x3 + x

x3 x2 + 1 x= 1

Rt may mn ta tm c x = 1. Thay x = 1 li h ta c 2 (y2 y+ 1) = 0y2 y+ 1 = 0 2.P T(2) P T(1)s c nhn tx 1

C th l (x 1) (y2 (x + 3) y+x2 x 2) = 0TH1 :x= 1thay vo th v nghim

TH2: Kt hp thm vi PT(1) ta c h mi : y2 (x + 3) y+x2 x 2 = 0 (3)x3 +y2 x2y+ x + xy2 y = 0



NguynM

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2.1 Cu 1 n cu 30 19

Nhn xt h ny c c im ging vi h ban u l bc y nh nhau. Vy ta li vit li htheo n y v hi vng n s li ng vi x no . Tht vy, l x =1

2. Tip tc thay n

vo h v ta s rt ra :

2P T(2) P T(1) (2x + 1) y2 (x 1) y+ x2 x + 2TH1 : x= 1

2 y=5 3

5

4TH2 : Kt hp vi (3) ta c

y2 (x 1) y+x2 x + 2 = 0y2 (x + 3) +x2 x 2 = 0

Vi h ny ta ch vic tr cho nhau s ra y= 1 x2 + 2 = 0(V nghim)Vy h cho c nghim :(x; y) =

1

2

;5 + 3

5

4 ,

1

2

;5 35

4

Cu 24

2 (x +y)(25 xy) = 4x2 + 17y2 + 105x2 + y2 + 2x 2y= 7

Gii

Hnh thc bi h c v kh ging vi cu 23

Mt cht nh gi v h ny- Cc bin x v y khng c lp vi nhau- Bc cao nht ca x 2 phng trnh nh nhau , y cng vyVi cc c im ny ta th vit h thnh 2 phng trnh theo n x v y v xem liu h cng vi x hoc y no khng. Cch lm vn nh cu 23. Vit theo x ta s khng tm c y,nhng vit theo y ta s tm c x = 2 khin h lun ng. Thay x = 2 vo h ta c

21y2 42y+ 21 = 0y2 2y+ 1 = 0 P T(1) 21P T(2) (x 2)

2y2 + 2xy+ 4y 17x 126 = 0

TH1 : x= 2 y= 1

TH2 :

2y

2

+ 2xy+ 4y 17x 126 = 0x2 + y2 + 2x 2y 7 = 0H ny c cch gii ri nh ??3.P T(2) P T(1) (x y+ 5)2 + 2x2 + x + 80 = 0(V nghim)Vy h cho c nghim : (x; y) = (2; 1)

Tip theo chng ta s n vi cu VMO 2004.



NguynM

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n


Cu 25

x3 + 3xy2 = 49x2 8xy+y2 = 8y 17x

Gii

Li gii ngn gn nht ca bi trn l :

P T(1) + 3.P T(2) (x + 1) (x + 1)2 + 3(y 4)2 = 0n y d dng tm ra nghim (x; y) = (1;4), (1;4)

Cu hi c t ra l bi ny tm hng s nh th no ? C rt nhiu cch gii thch nhngti xin trnh by cch gii thch ca ti :tuzki:Lm tng t theo nh hai cu 23 v 24 xem no. Vit li h cho thnh

3xy2 + x3 + 49 = 0y2 + 8(x + 1)y+ x2 17x= 0

Mt cch trc gic ta th vi x= 1. V sao ? V vi x= 1phng trnh 2 s khng cnphn y v c v 2 phng trnh s tng ng. Khi thay x= 1h cho tr thnh

3y2 + 48 = 0y2 16 = 0

Hai phng trnh ny tng ng. Tri thng ri !! Vy x =1chnh l 1 nghim cah v t h th hai ta suy ra ngay phi lm l P T(1) + 3.P T(2). Vic cn li ch l phntch nt thnh nhn t.

Tip theo y chng ta s n vi mt chm h d bn ca tng trn. Ti khng trnhby chi tit m ch gi v kt qu

Cu 26

y3 + 3xy2 = 28x2 6xy+y2 = 6x 10y

Gi : P T(1) + 3.P T(2) (y+ 1) (3(x 3)2 + (y+ 1)2) = 0Nghim ca h : (x; y) = (3;1), (3;1)

Cu 27

6x2y+ 2y3 + 35 = 05x2 + 5y2 + 2xy+ 5x + 13y= 0

Gi : P T(1) + 3.P T(2) (2y+ 5)

3

x +

1

2

2+

y+

5

2

2= 0



NguynM

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2.1 Cu 1 n cu 30 21

Cu 28

x3 + 5xy2 = 352x2 5xy 5y2 + x + 10y 35 = 0

Gi : P T(1) + 2.P T(2) (x 2) (5(y 1)2

+ (x + 3)2

) = 0

Cu 29

x3 + 3xy2 = 6xy 3x 49x2 8xy+y2 = 10y 25x 9

Gi : P T(1) + 3.P T(2) (x + 1) ((x + 1)2 + 3(y 5)2) = 0

im qua cc cu t cu 23 n cu 29 ta thy dng nh nhng cu h ny kh c bit.Phi c bit th nhng h s kia mi t l v ta tm c x = hay y = l nghim cah. Th vi nhng bi h khng c c may mn nh kia th ta s lm nh no. Ti xin giithiu mt phng php UCT rt mnh. C th p dng rt tt gii nhiu bi h hu t (kc nhng v d trn). l phng php Tm quan h tuyn tnh gia x v y. V ta skhng ch nhn hng s vo mt phng trnh m thm ch nhn c mt hm f(x)hay g(y)vo n. Ti s a ra vi v d c th sau y :

Cu 30

3x2 + xy 9x y2 9y= 02x3 20x x2y 20y= 0

Gii

Bi ny nu th nh cu 23, 24, 25 u khng tm ra ni x hay y bng bao nhiu l nghim cah. Vy phi dng php dng quan h tuyn tnh gia x v y. Quan h ny c th xy dng

bng hai cch thng dng sau :- Tm ti thiu hai cp nghim ca h- S dng nh l v nghim ca phng trnh hu t

Trc ht ti xin pht biu li nh l v nghim ca phng trnh hu t :Xt a thc : P(x) =anx

n + an1xn1 + ....+a1x + a0a thc c nghim hu t

p

q p l c ca a0 cn q l c ca an

OK ri ch ? By gi ta hy th xy dng quan h theo cch u tin, l tm ti thiu haicp nghim ca h ( Casio ln ting :v )D thy h trn c cp nghim l (0;0v(2;1)Chn hai nghim ny ln lt ng vi ta 2 im, khi phng trnh ng thng quachng s l : x + 2y= 0 x= 2y



NguynM

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Nh vy quan h tuyn tnh y l x= 2y. Thay li vo h ta c 9y (y+ 1) = 020y (y+ 1) (y 1) = 0

Sau ta chn biu thc ph hp nht nhn vo 2 phng trnh. y s l 20 (y 1) .P T(1) + 9.P T(2)Nh vy

20 (y 1) .P T(1) + 9.P T(2) (x + 2y) 18x2 + 15xy 60x 10y2 80y = 0TH1 : x= 2ythay vo (1)TH2 : Kt hp thm vi PT(1) na thnh mt h gm hai tam thc bit cch giiNghim ca h :

(x; y) = (0; 0), (2;

1), (10; 15),

15 1452

; 11

145 ,

15 +

145

2 ; 11 +

145

S dng cch ny chng ta thy, mt h phng trnh hu t ch cn tm c mt cpnghim l ta xy dng c quan h tuyn tnh v gii quyt bi ton. y chnh l uim ca n. Bn c th vn dng n vo gii nhng v d t 23 n 29 xem. Ti th lmcu 25 nh : Cp nghim l(1;4), (1;4)nn quan h xy dng y l x = 1. Thay livo h v ta c hng chn h s nhn.

Tuy nhin cch ny s chu cht vi nhng bi h ch c mt cp nghim hoc nghim qul khng th d bng Casio c. y l nhc im ln nht ca n

No by gi hy th xy dng quan h bng nh l nh.

Vi h ny v phng trnh di ang c bc cao hn trn nn ta s nhn avo phng trnhtrn ri cng vi phng trnh di. V bc ca x ang cao hn nn ta vit li biu thc saukhi thu gn di dng mt phng trnh bin x. C th l

2x3 + (3a y) x2 + (ay 9a 20) x y (ay+ 9a + 20) = 0()Nghim ca (*) theo nh l s l mt trong cc gi tr

1,12

,y2

,y, ....Tt nhin khng th c nghim x= 1

2hayx = 1c. Hy th vi hai trng hp cn li.

* Vi x= y thay vo h ta c 3y2 18y= 0y3 40y= 0

Khi ta s phi ly(y2 40).P T(1) 3(y 6).P T(2). R rng l qu phc tp. Loi ci ny.* Vi x= ythay vo h ta c

y2 = 03y3 = 0

Khi ta s ly 3y.PT(1) +P T(2). Qu n gin ri. Khi biu thc s l

(x + y)

2x2 + 6xy 3y2 + 27y+ 20 = 0Cch s hai rt tt thay th cch 1 trong trng hp khng tm ni cp nghim. Tuy nhinyu im ca n l khng phi h no dng nh l cng tm c nghim. Ta phi bit kt

hp nhun nhuyn hai cch vi nhau. V hy th dng cch 2 lm cc cu t 23 n 29 xem.N s ra nghim l hng s.

Lm mt cu tng t na. Ti nu lun hng gii.



NguynM

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2.2 Cu 31 n cu 60 23

2.2 Cu 31 n cu 60

Cu 31

x2y2 + 3x + 3y 3 = 0x2y 4xy 3y2 + 2y x + 1 = 0

Gii

P T(1) (y 1).P T(2) (x + y 1) 3y2 +xy 2y+ 2 = 0TH1 : x= 1 y. No problem !!!Th2 :

3y2 + xy 2y+ 2 = 0x2y 4xy 3y2 + 2y x + 1 = 0

y li l h c bit, ta tm c x= 3l nghim ca h. Thay vo v rt ra kt qu

PT(1) + PT(2) (x 3) (xy 1) = 0

Vy h cho c nghim (x; y) = (0; 1), (1; 0)

Bi vit v phng php UCT hay cn gi l h s bt nh kt thc y. Qua hn chccu ta thy : s dng phng php UCT nng cao (tm quan h tuyn tnh gia cc n) lmt phng php rt mnh v rt tt gii quyt nhanh gn cc h phng trnh hu t. Tuynhin nhc im ca n trong qu trnh lm l kh nhiu. Th nht : tnh ton qu tru b

v hi no. Hin nhin ri, dng quan h tuyn tnh kh, sau cn phi nhc cng phntch mt a thc hn n thnh nhn t. Th hai, nu s dng n mt cch thi qu s khinbn thn tr nn thc dng, my mc, khng chu my m suy ngh m c nhn thy l laou vo UCT, c khc g lao u vo khng ?

Mt cu hi t ra. Liu UCT c nn s dng trong cc k thi, kim tra hay khng ? Xintha, trong nhng VMO, cng lm tng ca h l dng UCT dng c bn, tc l nhnhng s thi. UCT dng c bn th ti khng ni lm g ch UCT dng nng cao th tt nhtkhng nn xi trong cc k thi. Th nht mt rt nhiu thi gian v sc lc. Th hai gy khkhn v c ch cho ngi chm, h hon ton c th gch b ton b mc d c th bn lmng. Vy nn : CNG NG LM RI MI DNG NH !! :D

y c l l bi vit ln nht m ti km vo trong cun sch. Trong nhng cu tip theoti s ci nhng bi vit nh hn vo. n xem nh. Nhng cu tip theo c th cn mt scu s dng phng php UCT. Vy nn nu thc mc c quay tr li t cu 20 m xem. Tmthi gc li , ta tip tc n vi nhng cu tip theo.



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Cu 32

x5 + y5 = 1x9 + y9 =x4 + y4

Gii

Nhn thy r rng y l loi h bn ng cp. Ta nhn cho hai v vi nhau c

x9 + y9 = (x4 + y4)(x5 + y5) x4y4(x + y) = 0

TH1 : x= 0 y= 1TH2 : y= 0 x= 1TH3 : x= ythay vo (1) r rng v nghimVy h cho c nghim (x; y) = (1; 0), (0; 1)

Cu 33

x3 + 2xy2 = 12y8y2 +x2 = 12

Gii

Li thm mt h cng loi, nhn cho hai v cho nhau ta c

x3 + 2xy2 =y(8y2 +x2) x= 2y

Khi (2) s tng ng12y2 = 12 y= 1, x= 2

Vy h cho c nghim (x; y) = (2; 1), (2;1)

Cu 34

x2 +y2 +

2xy

x + y = 1

x + y = x2 y

Gii

iu kin : x + y >0R rng khng lm n c t phng trnh (2). Th bin i phng trnh (1) xem

(1) (x + y)2 1 + 2xyx +y

2xy= 0

(x + y+ 1)(x + y 1) 2xy(x + y 1)x + y

= 0

C nhn t chung ri. Vi x + y= 1thay vo (2) ta c

1 = (1 y)2 y y= 0, y = 3



NguynM

inhTu

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2.2 Cu 31 n cu 60 25

Gi ta xt trng hp cn li. l x + y+ 1 = 2xy

x +y

x +y+ 1 = 1 x2 y2 x2 + y2 + x + y= 0

R rng sai v t iu kin cho ngay x +y >0Vy h cho c nghim (x; y) = (1; 0), (2;3)

Cu 35

x3 y3 = 3(x y2) + 2x2 +

1 x2 3

2y y2 + 2 = 0

Gii

iu kin :1 x 1, 0 y 2Thng th bi ny ngi ta s lm nh sau. phng trnh (1) mt cht

(1) x3 3x= (y 1)3 3(y 1)

Xt f(t) =t3 3tvi1 t 1th f(t) = 3t2 3 0Suy ra f(t)n iu v t suy ra x= y 1thay vo (2)Cch ny n. Tuy nhin thay vo lm vn cha phi l nhanh. Hy xem mt cch khc rt mim m ti lm

(2) x2 +

1 x2 + 2 = 3

2y y2 f(x) =g(y)

Xt f(x)trn min [1;1]ta s tm c 3 f(x) 134Ta li c : g(y) = 3

y(2 y) 3 y+ 2 y

2 = 3

Vyf(x) g(y). Du bng xy ra khi y= 1x= 1, x= 0 Thay vo phng trnh u ch c cp(x; y) = (0; 1)l tha mn


Cu 36

x3 3x= y3 3yx6 + y6 = 1

Gii

D thy phng trnh (1) cn xt hm ri, tuy nhin f(t) =t33tli khng n iu, cn phib thm iu kin. Ta s dng phng trnh (2) c iu kin. T (2) d thy 1 x, y 1.Vi iu kin r rng f(t)n iu gim v suy ra c x= yThay vo (2) ta c

2x6

= 1 x

= 1

62Vy h cho c nghim :(x; y) =

16

2;

16

2

,

1

6

2; 1

6

2



NguynM

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Cu 37

x3(2 + 3y) = 1x(y3 2) = 3

Gii

Nhn thy x= 0khng l nghim. H cho tng ng

3y+ 1 = 1

x33

x+ 2 =y3

y = 1x

Thay li (1) ta c

2x3 + 3x2 1 = 0

x= 1 y= 1x=

1

2y = 2

Vy h cho c nghim :(x; y) = (1;1),

1

2; 2

Cu 38

x2 + y2 + xy+ 1 = 4yy(x +y)2 = 2x2 + 7y+ 2

Gii

S dng UCT s thy y= 0l nghim ca h. Thay li v ta s c

2P T(1) +P T(2) y(x + y+ 5)(x +y 3) = 0 y= 0x= 5 y

x= 3 y

Vi y= 0thay li v nghimVi x= 5 ykhi phng trnh (1) s tng ng

(y+ 5)2 +y2

y2

5y+ 1 = 4y

V L

Tng t vi x= 3 ycng v nghimVy h cho v nghim



NguynM

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2.2 Cu 31 n cu 60 27

Cu 39

x +y x y= y2

x2 y2 = 9

Gii

iu kin : y min{x}Ta khng nn t n tng hiu v vn cn st li

y

2s lm bi ton kh khn hn. Mt cch

trc gic ta bnh phng (1) ln. T (1) ta suy ra

2x 2

x2 y2 = y2

4

n y nhn thy

x2 y2 theo (2) bng 3. Vy suy ra

2x 6 = y2

4 y2

= 8x 24Thay vo (2) ta c

x2 8x + 15 = 0 x= 3 y= 0(T M)x= 5 y= 4(T M)

x= 5 y= 4(T M)Vy h cho c nghim (x; y) = (3; 0), (5; 4), (5;4)

Cu 40

x y+ 1 =52

y+ 2(x 3)x + 1 = 34

Gii

iu kin : x, y 1Khng tm c mi quan h c th no. Tm thi ta t n d nhnt

x + 1 =a

0,

y+ 1 =b

0. H cho tng ng

a2 1 b= 5

2

b2 1 + 2a(a2 4) = 34

Ta th b=7

2 a2 t (1) vo (2) v c :

72 a

22+ 2a(a

2

4) 1

4 = 0

a= 3 b= 112

(L)

a=

2

b=

1

2(L)

a= 1 b= 52

(T M)

a= 2 b= 12

(L)

x= 0

y = 34



NguynM

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0;34

Cu 41

(x2 + xy+ y2)

x2 + y2 = 185

(x2 xy+ y2)

x2 + y2 = 65

Gii

Thot nhn qua th thy y l mt h ng cp bc 3 r rng. Tuy nhin nu tinh ta emcng 2 phng trnh cho nhau s ch cn li x2 + y2Cng 2 phng trnh cho nhau ta c

2(x2 +y2)

x2 +y2 = 250 x2 + y2 = 5Khi thay li h ta c

(25 +xy).5 = 185(25 xy).5 = 65

xy= 12x2 + y2 = 25

x= 3, y = 4x= 4, y = 3x= 3, y= 4x= 4, y= 3

Vy h cho c nghim (x; y) = (3; 4), (4; 3), (3;4), (4;3)

Cu 42

y

x+

x

y =

7xy

+ 1

x

xy+y

xy= 78

Gii

iu kin : xy 0H cho tng ng

x + yxy

=7 +

xy

xyxy(x +y) = 78

t x + y= a,xy=b. H cho tng ng

a b= 7ab= 78

a= 13b= 6

a= 6b= 13 (L)

x +y = 13xy= 36

x= 9, y= 4x= 4, y= 9

Vy h cho c nghim (x; y) = (9; 4), (4; 9)



NguynM

inhTu

n

2.2 Cu 31 n cu 60 29

Cu 43

x3 y3 = 9x2 + 2y2 x + 4y= 0

Gii

Dng UCTP T(1) 3.P T(3) (x 1)3 = (y+ 2)3 x= y + 3

n y d dng tm nghim (x; y) = (1;2), (2;1)

Cu 44

8x3y3 + 27 = 18y3

4x2y+ 6x= y2

Gii

y l mt h hay. Ta hy tm cch loi b 18y3 i. V y = 0khng l nghim nn (2) tngng

72x2y2 + 108xy= 18y3

n y tng r rng ri ch ? Th 18y3 t (1) xung v ta thu c

8x3y3

72x2y2

108xy+ 27 = 0

xy= 3

2

xy=

21

9

5

4

xy=21 + 9

5

4

Thay vo (1) ta s tm c y v x

y= 0(L)

y= 3

8(xy)3 + 27

18 = 3

2

5 3 x= 1

4

35

y= 3

8(xy)3 + 27

18 =

3

2 3 +

5x=

1

4 3 +

5Vy h cho c nghim :(x; y) =

1

4

35 ;3

2

5 3 ,1

4

3 +

5

;3

2

3 +

5



NguynM

inhTu

n


Cu 45

(x +y)

1 +

1

xy

= 5

(x2 + y2)

1 +

1

x2y2

= 9

Gii

iu kin : xy= 0Ta c nhn ra . H tng ng

x +y+1

x+

1

y = 5

x2 + y2 + 1

x2+

1

y2 = 9

x +

1

x

+

y+

1

y

= 5

x +1

x

2+

y+

1

y

2= 13

x +

1x

= 2, y+1y

= 3

x +1

x= 3, y+

1

y = 2

x= 1, y =

3 52

x=3 5

2 , y= 1


1;

3 52

,

35

2 ; 1

Cu 46

x2 + y2 + x + y = 18x(x + 1)y(y+ 1) = 72

Gii

Mt bi t n tng tch cng kh n gint x2 + x= a, y2 + y= b. Ta c

a + b= 18ab= 72

a= 12, b= 6

a= 6, b= 12

x2 + x= 6y2 +y = 12

x

2

+ x= 12y2 +y = 6

x= 2, x= 3y = 3, y = 4

x= 3, x= 4y = 2, y = 3

Vy h cho c c thy 8 nghim



NguynM

inhTu

n

2.2 Cu 31 n cu 60 31

Cu 47

x3 + 4y= y3 + 16x1 +y2 = 5(1 + x2)

Gii

H cho tng ng x3 16x= y (y2 4)y2 4 = 5x2

Nh vy phng trnh (1) s l

x3 16x= 5x2y x= 0, y = 2

y=x2 16

5x

Trng hp 2 thay vo (2) s l(x2 16)2

25x2 4 = 5x2

x2 = 1

x2 = 6431

x= 1, y = 3x= 1, y= 3

Vy h cho c nghim (x; y) = (0; 2), (0;2), (1;3), (1;3)

Cu 48 x +y2 x2 = 12

y

x

y2 x2 = 12

Gii

iu kin : y2 x2 x

y2 x2 sinh ra t vic ta bnh phng (1). Vy th bm theo hng xem. T (1)

ta suy tax2 + y2 x2 + 2x

y2 x2 = (12 y)2

y2 + 24 = (12 y)2 y= 5

Thay vo (2) ta c x

25 x2 = 12 x= 3, x= 4i chiu li thy tha mnVy h cho c nghim (x; y) = (3; 5), (4; 5)



NguynM

inhTu

n


Cu 49

x4 4x2 + y2 6y+ 9 = 0x2y+x2 + 2y 22 = 0

Gii

nu t x2 = ath h cho bin thnh h tam thc bc 2 ta hon ton bit cchgii. C th y s l

P T(1) + 2.P T(2) (x2 + y)2 2(x2 + y) 35 = 0

TH1 : x2 +y = 7 x2 = 7 ythay (2) ta c

(7 y)y+ 7 y+ 2y 22 = 0

y = 3 x= 2y = 5 x= 2

TH2 : x2

+y = 5 x2

= 5 y. Hon ton tng t thay (2) s cho y v nghimVy h cho c nghim : (x; y) = (2; 3), (2;3), (2;5), (2;5)

Cu 50

x2 +y+ x3y+ xy+ y2x= 54

x4 +y2 + xy(1 + 2x) = 54

Giiy l cu Tuyn sinh khi A - 2008. Mt cch t nhin khi gp hnh thc ny l ta tin hnhnhm cc s hng liH cho tng ng

(x2 + y) +xy+ (x2 +y)xy= 5

4

(x2 + y)2 +xy = 54

n y hng i r rng. t x2 +y = a, xy= b ta c

a + b + ab= 54

a2 + b= 54

a= 0, b=

5

4

a= 12

, b= 32

x2 + y= 0xy= 5

4

x2 +y = 12

xy= 32

x= 3

5

4, y= 3

25

16

x= 1, y = 32

Vy h cho c nghim (x; y) =

3

5

4; 3

25

16

,

1;3

2



NguynM

inhTu

n

2.2 Cu 31 n cu 60 33

Cu 51

x2 + 1 + y(y+x) = 4y(x2 + 1)(x + y 2) =y

Gii

H gn nh ch l cu chuyn ca x2 + 1v x+y. Tuy nhin ychen vo khin h tr nnkh chu. Hy dit yi . Cch tt nht l chia khi m y = 0khng phi l nghim cah. H cho tng ng

x2 + 1

y + x + y 2 = 2

x2 + 1

y (x + y 2) = 1

Hng i r rng. t x2 + 1

y =a, x +y 2 =b

H cho tr thnh a +b= 2ab= 1

a= 1b= 1

x2 + 1 =yx + y= 3

x= 1, y= 2x= 2, y = 5

Vy h cho c nghim (x; y) = (1; 2), (2;5)

Cu 52 y+ xy2 = 6x2

1 +x2y2 = 5x2

Gii

Loi h ny khng kh. tng ta s chia bin v phi tr thnh hng sNhn thy x= 0khng l nghim. H cho tng ng

y

x2+

y2

x = 6

1

x2+y2 = 5

y

x

1

x+y

= 6

1

x+ y

2

2 yx

= 5

t y

x=a,

1

x+ y=b. H tr thnh

ab= 6b2 2a= 5

a= 2b= 3

y = 2x1

x+ y= 3

x= 1, y = 2

x=1

2, y = 1

Vy h cho c nghim (x; y) = (1; 2),

1

2; 1



NguynM

inhTu

n


Cu 53

x2 + 2y2 =xy + 2y2x3 + 3xy2 = 2y2 + 3x2y

Gii

mt cht y l h bn ng cp. Nu ta vit li nh sau x2 + 2y2 xy= 2y2x3 + 3xy2 3x2y= 2y2

T ta c

2y2(x2 + 2y2 xy) = 2y 2x3 + 3xy2 3x2y 4y (y x) x2 xy+y2 = 0TH1 : y= 0 x= 0

TH2 : x= y = 0TH3 : x= y thay vo (1) ta c

2y2 = 2y

x= y = 0x= y = 1

Vy h cho c nghim (x; y) = (0; 0), (1; 1)

Cu 54 2x2y+y3 = 2x4 + x6

(x + 2)y+ 1 = (x + 1)2

Gii

iu kin : y 1Khai thc t (1). C v nh l hm no . Chn chia cho ph hp ta s c mc ch, ys chia cho x3 v x= 0khng l nghim ca h. PT(1) khi s l

2y

x+ y

x3

= 2x + x3 yx

=x y = x2

Thay vo (2) ta s c

(x + 2)

x2 + 1 = (x + 1)2 (x + 2)2 x2 + 1 = (x + 1)4 x= 3, y= 3(T M)x= 3, y = 3(T M)

Vy h cho c nghim : (x; y) = (3;3)

Ta s n mt cu tng t n



NguynM

inhTu

n

2.2 Cu 31 n cu 60 35

Cu 55

x5 + xy4 =y10 +y64x + 5 +

y2 + 8 = 6

Gii

iu kin : x 54

Thy y= 0khng l nghim ca h. Chia 2 v ca (1) cho y5 ta c

x

y

5+

x

y =y5 + y x

y =y x= y2

Thay vo (2) ta c

4x + 5 +

x + 8 = 6 x= 1 y = 1

Vy h cho c nghim (x; y) = (1;1)

Cu 56

xy+ x + 1 = 7yx2y2 + xy+ 1 = 13y2

Gii

y l cu Tuyn sinh khi B - 2009. Cc gii thng thng nht l chia (1) cho y , chia (2)choy2 sau khi kim tra y= 0khng phi l nghim. Ta s c

x +x

y+

1

y= 7

x2 +x

y+

1

y2 = 13

x +1

y+

x

y = 7

x +1

y

2 x

y = 13

a + b= 7a2 b= 13

a= 4, b= 3a= 5, b= 12

x +1

y = 4

x= 3y x +1y = 5x= 12y

x= 1, y=

1

3x= 3, y= 1


1;1

3

, (3; 1)

Tip tc ta n thm mt cu tuyn sinh na



NguynM

inhTu

n


Cu 57

x4 + 2x3y+ x2y2 = 2x + 9x2 + 2xy= 6x + 6

Gii

tht k nu ta th kho lo xyln (1) s ch cn li phng trnh n x. D s l bc 4nhng liu th n nhiu. H vit li

x4 + 2x2(xy) +x2y2 = 2x + 9

xy=6x + 6 x2

2

T (1) s tng ng

x4

+ x2

(6x + 6 x2

) +6x + 6 x

2

22

= 2x + 9 x= 4x= 0 y=17

4V L

Vy h cho c nghim (x; y) =4;17

4

Cu 58

3

1 +x +

1 y = 2x2 y4 + 9y= x(9 +y y3)

Gii

iu kin : y 1Khng lm n g c t (1). Xt (2). 1 to th (2) c th phn tch c thnh

(x y) (9 x y3) = 0

x= yx= 9 y3

Vi x= y thay vo (1) ta s c

3

1 +y+

1 y= 2 a + b= 2a3 +b2 = 2

b 0 a= 1, b= 1a= 1 3, b= 3 + 3

a=

3 1, b= 3 3 y= 0y= 63 11

y= 63 11

Vi x= 9 y3 thay vo (1) ta s c3

10 y3 +

1 y = 2

Ta c3

10 y3 +

1 y 3

9> 2

Vy h cho c nghim : (x; y) = (0; 0), (63 11;63 11), (63 11;63 11)



NguynM

inhTu

n

2.2 Cu 31 n cu 60 37

Cu 59

xy+

1 y= y2

y

x 1y = 1

Gii

iu kin : x 1, 0 y 1Thot nhn bi ton ta thy nh lc vo m cung nhng cn thc. Tuy nhin ch vi nhngnh gi kh n gin ta c th chm p bi tonVit li phng trnh (2) nh sau

2

y

x 1 = y 1

T iu kin d thy V T 0 V PDu bng xy ra khi x= y = 1Vy h cho c nghim (x; y) = (1; 1)

Cu 60

x

17 4x2 +y

19 9y2 = 317 4x2 +

19 9y2 = 10 2x 3y

Gii

iu kin :172 x

172

,193 y

193

Bi ton ny xut hin trn thi th ln 2 page Yu Ton hc v ti l tc gi ca n. tng ca n kh n gin, ph hp vi 1 thi tuyn sinh x

17 4x2 lin quan n 2xv17 4x2, y

19 9y2 lin quan n 3yv19 9y2.

V tng bnh phng ca chng l nhng hng s. y l c s ta t nt 2x +

17 4x2 =a, 3x +

19 9y2 =b. H cho tng ng

a + b= 10a2 17

4 +

b2 196

= 3

a= 5, b= 5a= 3, b= 7

TH1 :

2x +17 4x2 = 53y+

19 9y2 = 5

x= 1

2x= 2

y=5 13

6

TH2 :

2x +

17 4x2 = 33y+

19 9y2 = 7 (Loi)

Vy h cho c nghim :(x; y) =

1

2;5 +

13

6

1

2;5 13

6

2;

5 +

13

6

2;

5 136

V y l tng gc ca n. Hnh thc n gin hn mt cht



NguynM

inhTu

n


2.3 Cu 61 n cu 90

Cu 61

x

5 x2 +y

5 4y2 = 15 x2 +5 4y

2 =x 2y

Nghim : (x; y) = (1;1),

2;12

Cu 62

x3 xy2 + y3 = 14x4 y4 = 4x y

Gii

R rng l mt h a v c dng ng cp bng cch nhn cho v vi v. Tuy nhin, biny nu s dng php th tt ta s a v mt kt qu kh p mtPhng trnh (2) tng ng

4x(x3 1) =y(y3 1)n y ta rt x3 1vy3 1t (1). C th t (1) ta c

x3 1 =y3 y2xy3 1 =xy2 x3

Thay tt c xung (2) v ta thu c

4xy2(y x) = xy(x2 y2)

x= 0y = 0x= y4y= y + x

y= 1x= 1x= y = 1

y= 13

25, x=

33

25

Vy h cho c nghim (x; y) = (0; 1), (1; 0), (1; 1),

13

25;

33

25

Cu 63

x +

x2 y2x

x2 y2 +

xx2 y2x +

x2 y2 =

17

4

x(x +y) +

x2 + xy+ 4 = 52

Gii

iu kin : x =

x2 y2, x2 y2 0, x2 +xy+ 4 0Hnh thc bi h c v kh khng b nhng nhng tng th l ht. Ta c th khai thc c2 phng trnh. Pt(1) c nhiu cch x l : ng cp, t n, lin hp. Ti s x l theo hngs 3. (1) khi s l

x +

x2 y22

x2 (x2 y2) +

x

x2 y22

x2 (x2 y2) =17

4 2 (2x

2 y2)y2

=17

4 y= 4x

5



NguynM

inhTu

n

2.3 Cu 61 n cu 90 39

Tip tc khai thc (2). D thy t

x2 +xy+ 4 =t 0th (2) tr thnh

t2 + t= 56

t= 7t= 8(L) x

2 +xy = 45

Kt hp li ta c y= 4

5x

x2 + xy= 45

x= 5, y = 4x= 5, y= 4x= 15, y= 12x= 15, y= 12

Vy h cho c nghim : (x; y) = (5;4), (5; 4), (15; 12), (15;12)

Cu 64x +y+x

y = 2

y+ xy x= 1Gii

iu kin : x, y 0,y min{x},x min{y}Khng tm c mi lin h g t c hai phng trnh, ta tin hnh bnh phng nhiu ln ph v ton b cn thc kh chu. Phng trnh (1) tng ng

2x + 2

x2 y= 4

x2 y= 2 x x2 y= x2 4x 4 4x y= 4Lm tng t phng trnh (2) ta s c : 4x

4y =

1. Kt hp 2 kt qu li d dng tm

c x,yVy h cho c nghim : (x; y) =

17

12;5

3

Cu 65

x + 2xy

3

x2 2x + 9 =x2 + y

y+ 2xy

3y2 2y+ 9=y2 +x

Gii

Hnh thc ca bi h l i xng. Tuy nhin biu thc kh cng knh v li nhn xt thyx= y = 1l nghim ca h. C l s nh giCng 2 phng trnh li ta c

x2 + y2 = 2xy

1

3

x2 2x + 9 + 1

3

y2 2y+ 9

T ta nhn xt c nghim th xy 0v l 3

t2 2t+ 9 2nn ta nh gix2 + y2 2xy

1

2+

1

2

(x y)2 0



NguynM

inhTu

n


Du bng xy ra khi (x; y) = (1; 1)

Cu 66

6

x

y 2 = 3x y+ 3y2

3x +

3x y= 6x + 3y 4

Gii

iu kin : y= 0, 3x y, 3x +3x y 0Phng trnh (1) khi s tng ng

6x

2y=y3x y+ 3y2

2 (3x

y)

y3x y 3y

2 = 0

3x y = y

3x y =3y

2

TH1 :

3x y= y. T y suy ra y 0v3x= y2 + ythay tt c vo (2) ta c

2

y2 + y y= 2 y2 + y+ 3y 4 2y2 + 7y 4 = 0y 0 y = 4 x= 4

TH2 :

3x y=3y2

. T y suy ra y 0v3x= 9y2

4 + ythay tt c vo (2) ta cng s tm

c y =8

9x=

8

9Vy h cho c nghim (x; y) = (4;4),

8

9;8

9

Cu 67

(3 x)2 x 2y2y 1 = 03

x + 2 + 2

y+ 2 = 5

Gii

iu kin : x 2, y12

Phng trnh (1) tng ng

(2 x)2 x +2 x= (2y 1)

2y 1 +

2y 1 f(2x 1) =f(

2y 1)

Vi f(x) =x3 +xn iu tng. T suy ra

2 x= 2y 1 x= 3 2ythay vo (2)ta c

3

5 2y+ 2

y+ 2 = 5

a + 2b= 5a3 + 2b2 = 9

a= 1, b= 2

a=3 654

, b= 23 + 658

a=

65 3

4 , b=

23 658



NguynM

inhTu

n

2.3 Cu 61 n cu 90 41

y= 2

y=233 + 23

65

32

y=233 2365

32

Vy h cho c nghim

(x; y) = (1;2),

23

65 18516

;233 2365

32

23

65 + 185

16 ;

233 + 23

65

32

S dng tnh n iu ca hm s cng l mt hng kh ph bin trong gii h phng trnh.Ch cn kho lo nhn ra dng ca hm, ta c th rt ra nhng iu k diu t nhng phngtrnh khng tm thng cht no

Cu 681 +xy+ 1 +x + y= 2

x2y2 xy= x2 + y2 + x + y

Gii

iu kin : xy 1, x + y 1Mt cht bin i phng trnh (2) ta s c

x2y2 + xy= (x +y)2 + x + y (xy x y)(xy+ x + y+ 1) = 0

x +y =xyx +y = xy 1

TH1 : xy=x + ythay vo (1) ta c2

1 +xy = 2 xy= 0 x= y = 0TH2 : x + y = xy 1thay vo (1) ta c

1 +xy+xy= 2(V L)

Vy h cho c nghim : (x; y) = (0; 0)

Cu 69

x + 3x yx2 + y2

= 3

y x + 3yx2 +y2

= 0

Gii

Ti khng nhm th bi ton ny xut hin trn THTT, tuy nhn hnh thc ca h kh pmt v gn nh nhng khng h d gii mt cht no. Hng lm ti u ca bi ny l phcha. Da vo tng h kh i xng ng thi di mu nh l bnh phng ca Moun mta s dng cch ny. Hng gii nh sau

PT(1)+i.PT(2) ta s c

x + yi +3(x yi) (xi + y)

x2 + y2 = 0



NguynM

inhTu

n


t z=x +yikhi phng trnh tr thnh

z+3z iz|z|2 = 3 z+

3z izz.z

= 3 z+3 iz

= 3

z= 2 + iz= 1 i

Vy h cho c nghim (x; y) = (2; 1), (1;1)Hnh thc ca nhng bi h ny kh d nhn thy. Th lm mt s cu tng t nh.

Cu 70

x +5x + 7

5y

x2 +y2 = 7

y+7

5x 5yx2 + y2

= 0

Cu 71

x + 5x yx2 + y2

= 3

y x + 5yx2 +y2

= 0

Cu 72

x +16x 11y

x2 +y2 = 7

y 11x + 16yx2 + y2

= 0

Cu 73

(6 x)(x2 + y2) = 6x + 8y(3 y)(x2 +y2) = 8x 6y

Gi : Chuyn h cho v dng

x +6x + 8y

x2 +y2 = 6

y+8x 6yx2 + y2

= 3

Nghim : (x; y) = (0; 0), (2;1), (4; 2)

Phc ha l mt phng php kh hay gii h phng trnh mang tnh nh cao. Khngch vi loi h ny m trong cun sch ti s cn gii thiu mt vi cu h khc cng s dngphc ha kh p mt.



NguynM

inhTu

n

2.3 Cu 61 n cu 90 43

Cu 74

4x2y2 6xy 3y2 = 96x2y y2 9x= 0

Gii

y l mt bi ton cng kh p mt. Thy x= 1l nghim ca h . Ta suy raP T(1) +P T(2) (x 1)(4y2(x + 1) + 6xy 9) = 0

TH1 : x= 1 y= 3TH2 : 4y2(x + 1) + 6xy 9 = 0V x= 0khng l nghim. Suy ra 4y2x(x + 1) + 6x2y 9x= 0(*)V sao nhn xvo y. UCT chng ? Ti ch gii thiu cho cc bn UCT nng cao thi chti ch dng bao gi. L do ch n gin ti mun xut hin 6x2y 9x= y2 t (2) thiVy (*)

4y2x(x + 1) +y2 = 0

y2(2x + 1)2 = 0

TH1 : y= 0v nghimTH2 : x= 1

2 y= 3, y= 3

2

Vy h cho c nghim : (x; y) = (1; 3),1

2; 3

,

1

2;3

2

Cu 75

x2

(y+ 1)2+

y2

(x + 1)2 =

1

2

3xy=x + y+ 1

Gii

iu kin x, y= 1Bi ton ny c kh nhiu cch gii. Ti xin gii thiu cch p nht ca bi nyp dng Bt ng thc AMGMcho v tri ca (1) ta c

V T 2xy(x + 1)(y+ 1)

= 2xy

xy+ x + y+ 1=

2xy

xy+ 3xy =

1

2

Du bng xy ra khi (x; y) = (1; 1),13 ;13

Cu 76

3y2 + 1 + 2y(x + 1) = 4y

x2 + 2y+ 1y(y x) = 3 3y

Gii

iu kin : x2 + 2y+ 1

0

Khng lm n g c t (2). Th bin i (1) xem sao. PT(1) tng ng

4y2 4y

x2 + 2y+ 1 +x2 + 2y+ 1 =x2 2xy+ y2

2y

x2 + 2y+ 12

= (x y)2



NguynM

inhTu

n


x2 + 2y+ 1 = 3y xx2 + 2y+ 1 =x + y

C v hi o nh ? Nhng mt cht th (1) c vc dng ca cc hng ng thc nn tangh n hng ny

By gi x l hai trng hp kia th no ? Chc bnh phng thi. Tt qu ! Phng trnh sch cn li xyv ym nhng ci th (2) c cTH1 :

x2 + 2y+ 1 = 3y x

3y xx2 + 2y+ 1 = 9y2 6xy+ x2

3y x6xy= 9y2 2y 1xy= y2 + 3y 3(2)

x= 1, y= 1(T M)

x=415

51 , y=

17

3(T M)

TH2 :

x2 + 2y+ 1 =x +y

x +y 0x

2

+ 2y+ 1 =x

2

+ 2xy+ y

2

x + y 02xy=

y2 + 2y+ 1

xy= y2 + 3y 3 x= 1, y= 1

x=

41

21 , y = 7

3(L)

Vy h cho c nghim : (x; y) = (1; 1),

415

51;

17

3

Nh chng ta bit. Tam thc bc hai c kh nhiu ng dng trong gii ton v h cngkhng phi l ngoi l. Ch vi nhng nh gi kh n gin : t iu kin ca tamthc c nghim m ta c th tm ra cc tr ca cc n. T nh gi v gii quyt nhngbi ton m cc phng php thng thng cng b tay. Loi h s dng phng php ny

thng cho di hai dng chnh. Th nht : cho mt phng trnh l tam thc, mt phngtrnh l tng hoc tch ca hai hm f(x) v g(y). Th hai : cho c 2 phng trnh u lphng trnh bc hai ca 1 n no . Hy th lt qua mt chm h loi ny nh.

Cu 77

x4 + y2 =698

81x2 + y2 + xy 3x 4y+ 4 = 0

GiiHnh thc ca h : mt phng trnh l tam thc bc hai mt c dng f(x) +g(y)v mt skh khng b. Ta hy khai thc phng trnh (2) bng cch nh gi Vit li phng trnh (2) di dng sau

x2 + (y 3)x + (y 2)2 = 0()y2 + (x 4)y+ x2 3x + 4 = 0()

(*) c nghim th x 0 (y 3)2 4(y 1)2 0 1 y73

(**) c nghim th y 0 (x 4)4 4(x2 3x + 4) 0 0 x 43

T iu kin cht ca hai n gi ta xt (1) v c mt nh gi nh sau

x4 + y2

4

3

4+

7

3

2=

697

81 0 V Pnn v nghimVy h cho c nghim : (x; y) = (0; 1), (1;1)

Cu 86

x3(4y2 + 1) + 2(x2 + 1)

x= 6

x2y(2 + 2

4y2 + 1) =x +

x2 + 1

Gii

iu kin : x 0Hnh thc ca bi h r rng l kh rc ri. Tuy nhin, (2) nu ta chia c 2 v cho x2th s c lp c xvyv hi vng s ra c iu g.

Nhn thy x= 0khng l nghim. Chia 2 v ca (2) cho x2

ta c

2y+ 2y

4y2 + 1 = 1

x+

1

x

1

x2+ 1



NguynM

inhTu

n


R rng 2 v u c dng f(t) =t +t

t2 + 1v hm ny n iu tng. Vy t ta suy ra

c 2y= 1

xthay vo (1) ta c

x31x2 + 1

+ 2(

x2

+ 1)

x

= 6

x3 + x + 2(x2 + 1)x= 6R rng v tri n iu tng vi iu kin ca x. Vy x= 1l nghim duy nht


1;1

2

Cu 877x + y+ 2x + y = 52x + y+ x y= 2

Gii

y l cu trong VMO 2000-2001. Khng hn l mt cu qu khiu kin : y min{2x;7x}Xut hin hai cn thc vy th t

7x + y = a ,

2x +y = bxem

Nhng cn x yth th no ? Chc s lin quan n a2, b2. Vy ta s dng ng nht thc

x y = k(7x + y) +l(2x + y) k= 35 , l= 85Vy h cho tng ng

a + b= 5

b +3a2

5 8b

2

5 = 2

a, b 0

a=15 77

2

b=

77 5

2

7x + y=151 1577

2

2x + y=51 577

2

x= 10 77y =

11 772


10

77;

11

77

2

Mt cch khc cng kh tt. t

7x + y = a,

2x + y= bv ta xy dng mt h tm sau a + b= 5a2 b2 = 5x

a +b= 5a b= x b=

5 x2

Thay vo (2) v ta c5 x

2 + x y= 2 x= 2y 1

n y thay li vo (2) v ta cng ra kt qu

Mt v d tng t ca bi ny



NguynM

inhTu

n

2.3 Cu 61 n cu 90 49

Cu 88

11x y y x= 17

y x + 6y 26x= 3

Nghim : (x; y) = 37

20;81

10

Cu 89

3x

1 +

1

x + y

= 2

7y

1 1

x +y

= 4

2

Gii

y l cu trong VMO 1995-1996. Mt tng kh p mt m sng toiu kin : x, y 0, x + y >0H cho tng ng

1 + 1

x +y =

23x

1 1x + y

=4

27y

1

x + y =

13x2

2

7y

1 = 1

3x+

2

27y

1x + y

=

1

3x 2

2

7y

1

3x+

2

27y

1x + y

= 1

3x 8

7y 21xy= (x + y)(7y 3x)

(y 6x)(7y+ 4x) = 0 y= 6xThay vo phng trnh u ta c

1 + 1

7x=

23x

x= 11 + 4

7

21 y = 22

7 +

87

Mt cch khc c th s dng trong bi ny l phc ha. N mi xut hin gn ytx= a >0,y= b >0. Ta c h mi nh sau

a +

a

a2 + b2 =

23

b ba2 + b2

=4

27

P T(1) +i.P T(2) (a +bi) + a bia2 + b2

= 2

3+

4

27

i

t z=a +biphng trnh cho tr thnh

z+1

z =

23

+4

27

i z a, b x, y



NguynM

inhTu

n



11 + 4

7

21 ;

22

7 +

87

Bi h ny c kh nhiu d bn phong ph. Ti xin gii thiu cho cc bn

Cu 90

x

3

1 +

6

x + y

=

2

y

1 6

x +y

= 1

Nghim : (x; y) = (8; 4)

2.4 Cu 91 n cu 120

Cu 91

x

1 12

y+ 3x

= 2

y

1 +

12

y+ 3x

= 6

Nghim : (x; y) = (4 + 2

3; 12 + 6

3)

Cu 92

10x

1 +

3

5x + y

= 3

y

1 3

5x +y

= 1

Nghim : (x; y) =

2

5; 4

Cu 93

4

x

1

4+

2

x +

y

x +y

= 2

4

y

1

42

x +

y

x + y

= 1

Tip theo ta n mt vi v d v s dng phng php lng gic ha trong gii h phng trnh



NguynM

inhTu

n

2.4 Cu 91 n cu 120 51

Cu 94

x

1 y2 +y1 x2 = 1(1 x)(1 +y) = 2

Gii

iu kin :|x| 1,|y| 1iu kin ny cho ta tng lng gic ha. t x= sina, y= sinbvi a, b

2;

2

Phng trnh u tng ng

sinacosb + sinbcosa= 1 sin(a + b) = 1 a + b= 2

Phng trnh (2) tng ng

(1

sina)(1 +sinb) = 2

(1

sina)(1 +cosa) = 2

a=

2

a= 0 b=

b=

2

x= 1, y= 0(L)x= 0, y= 1

Vy h cho c nghim : (x; y) = (0; 1)

Cu 95 2y=x(1 y2)3x

x3 =y(1

3x2)

Gii

Thot nhn ta thy c v h ny cng xong, ch c g khi vit n di dng xy2 =x 2yx3 3x2y= 3x y

a n v dng ng cp, nhng ci chnh y l nghim n qu l. Vy th hng khcxem. Vit li h cho sau khi xt

x= 2y1 y2y =

3x x31 3x2

Nhn biu thc v phi c quen thuc khng ? Rt ging cng thc lng gic nhn i vnhn ba ca tan. Vy tng ny rat x= tanvi

2;

2

. T PT(2) ta s c

y=3tan tan3

1

3tan2

= tan 3

M nh th theo (1) ta s c

x= 2tan3

1 tan23 = tan 6



NguynM

inhTu

n


T suy ra

tan = tan 6 = k5 =

2

5 ;

5; 0;

5;2

5

Vy h cho c nghim : (x; y) = tan 25

;tan6

5 ,tan

5

;tan3

5 , (0; 0)

Lm mt bi tng t nh.

Cu 96

y =3x x31 3x2

x=3y y31 3y2

S dng phng php lng gic ha trong gii h phng trnh cn phi nm r cc hngng thc, ng thc, cng thc lng gic, v cn mt nhn quan tt pht hin mt biuthc no ging vi mt cng thc lng gic.

Cu 97 x3y(1 +y) +x2y2(2 +y) +xy3 30 = 0x2y+x(1 +y+y2) +y

11 = 0

Gii

y l mt h kh mnh nhng hay. Nhn vo 2 phng trnh ta thy cc bin "kt dnh" vinhau kh tt v hng s c v nh ch l k ng ngoi. Vy hy vt hng s sang mt bn vthc hin bin i v tri. H phng trnh cho tng ng

xy(x + y)(x +y+ xy) = 30xy(x + y) +x + y+ xy= 11

n y tng r rng. t a= xy(x + y), b= xy + x + yv h cho tng ng

ab= 30a + b= 11

a= 5, b= 6a= 6, b= 5

xy(x +y) = 5xy+ x + y= 6

xy(x +y) = 6xy+ x + y= 5

TH1 :

xy(x + y) = 6xy+ x + y= 5

xy= 2x + y = 3

xy= 3x + y = 2

(L)

x= 2, y= 1x= 1, y= 2

TH2 :

xy(x + y) = 5xy+ x + y= 6

xy= 5

x + y = 1 (L) xy= 1x + y = 5

x= 5 212 , y = 5 + 212

x=5 +

21

2 , y=

5212



NguynM

inhTu

n

2.4 Cu 91 n cu 120 53

Vy h cho c nghim : (x; y) = (1; 2), (2; 1),

521

2 ;

5 212

Tc gi ca n rt kho lo trn nhiu ln cch t n tng tch vo mt h, gy nhiu kh

khn cho ngi lm

Cu 98

sin2x +

1

sin2x+

cos2y+

1

cos2y =

20y

x + ysin2y+

1

sin2y+

cos2x +

1

cos2x=

20x

x + y

Gii

Bi ton xut hin trong VMO 2012-2013. Hnh thc bi h c s khc l khi c c hmlng gic chen chn vo. Vi kiu h ny nh gi l cch tt nhtTa s cng hai phng trnh vi nhau v s chng minh V T 210 V Pp dng Bt ng thc Cauchy Schwarzcho v phi ta c

20y

x + y+

20x

x + y

2

20y

x + y+

20x

x +y

= 2

10

Gi ta s chng minh : V T

2

10tc l phi chng minhsin2x +

1

sin2x+

cos2x +

1

cos2x

10

V T =

sin x 1

sin x

2+

22

+

cos x 1

cos x

2+

22

1

sin x+

1

cos x (sin x + cos x)

2+

2

22

Hin nhin ta c sinx + cosx

2nn1

sin x+

1

cos x (sin x + cos x) 4

sin x + cos x

2 42

2 =

2

VyV T 2 + 8 = 10. Tng t vi bin yv ta c iu phi chng minhng thc xy ra khi x= y =

4+ k2



NguynM

inhTu

n


Cu 99

x

xx= yy+ 8yx y= 5

Gii

iu kin : x, y 0 h ny cho mt phng trnh n gin qu. Th thng ln (1) chng ? Khng nn ! Bin i1 to ri hy th. Hng bin i kh n gin l lm ph v cn thcPhng trnh (1) tng ng

x(x 1) = y(y+ 8) x(x 1)2 =y(y+ 8)2

n y thc hin th x= y + 5 ln (1) v ta c

(y+ 5)(y+ 4)2 =y(y+ 8)2

y= 4

x= 9


Cu 100

1x

+y

x=

2

x

y + 2

y

x2 + 1 1 = 3x2 + 3Gii

iu kin : x >0, y= 0R rng vi iu kin ny th t (2) ta thy ngay c nghim th y >0Phng trnh (1) tng ng

x + y

x =

2 (

x +y)

y

x +y = 0(L)

y= 2x

Vi y= 2xthay vo (2) ta c

2xx2 + 1 1 = 3x2 + 3 2x3x2 + 1 = 2x x2 + 1 = 2x2x 3

R rng v tri n iu tng v v phi n iu gim nn phng trnh ny c nghim duynht x=

3 y = 23

Vy h cho c nghim (x; y) = (

3; 2

3)



NguynM

inhTu

n

2.4 Cu 91 n cu 120 55

Cu 101

y= x3 + 3x + 4x= 2y3 6y 2

Gii

Hnh thc bi h kh gn nh nhng cng khin nhiu ngi phi lng tng. Nhn xtx= y = 2l nghim. Ta tin hnh tch nh sau

y 2 = (x + 1)2(x 2)x 2 = (y+ 1)2(y 2)

n y nhn cho v vi v ta c

2(y 2)2(y+ 1)2 = (x + 1)2(x 2)2

D thy V T 0 V P. y ng thc xy ra khi x= y = 2

Cu 102

x3 xy2 + 2000y= 0y3 yx2 500x= 0

Gii

D dng a c v h ng cp. Nhng ta bin i mt to n ti u.

H cho tng ng

x (x2 y2) = 2000yy(x2 y2) = 500x 500x

2(x2 y2) = 2000y2(x2 y2)

x= yx= yx= 2yx= 2y

Thay li vi mi trng hp vo (1) v ta c

y = 0, x= 0

y = 1010

3 , x= 20103y = 10

10

3 , x= 20

10

3

Vy h cho c nghim : (x; y) = (0; 0),

20

10

3;10

10

3



NguynM

inhTu

n


Cu 103

3

x2 + y2 1+ 2y

x= 1

x2 +y2 + 4x

y = 22

Gii

tng t n ph r rng. t x2 + y2 1 =a , yx

=b . H cho tng ng

3

a+ 2b= 1

a +4

b = 21

a= 7, b=

2

7

a= 9, b=1

3

x2 +y2 = 82x= 7y

x2 +y2 = 10x= 3y

y= 4

2

53, x= 14

2

53x= 3, y = 1

Vy h cho c nghim : (x; y) = (3;1)14253 ;4253

Cu 104

x +1

y+

x + y 3 = 3

2x + y+1

y = 8

Giiiu kin : y= 0, x +1

y 0, x + y 3

tng t n ph cng kh r rng.

t

x +1

y =a 0,x + y 3 =b 0. H cho tng ng

a + b= 3a2 + b2 = 5

a= 1, b= 2

a= 2, b= 1

x +1

y = 1

x + y

3 = 4

x +1y

= 4

x + y 3 = 1

x= 410, y= 3 + 10x= 4 +

10, y = 3

10

x= 3, y= 1x= 5, y= 1

Vy h cho c nghim : (x; y) = (3; 1), (5;1)(4 10;310)



NguynM

inhTu

n

2.4 Cu 91 n cu 120 57

Cu 105

x3(2 + 3y) = 8x(y3 2) = 6

Gii

y l mt cu kh ging cu s 37Nghim : (x; y) = (2;1), (1;2)

Cu 106

2x2y+ 3xy= 4x2 + 9y7y+ 6 = 2x2 + 9x

Gii

Bi ny nu li ngh c th dng mn v th thn chng y vo PT(1). Nhng hy dng UCT y s tt hn.Nhn thy y = 3l nghim (ci ny gi li nh, ti khng gii thch na), thay y = 3vo hta c

2x2 + 9x 27 = 027 2x2 + 9x= 0

Nh vy hng ca ta s cng hai phng trnh ban u li v nhn ty 3s xut hin. Vy

P T(1) +P T(2) (3 y) 2x2 + 3x 2 = 0

n y d dng gii ra (x; y) =2;16

7

,

1

2;1

7

,

3(3

33)4

; 3

Cu 107

x2 + 3y= 9y4 + 4(2x 3)y2 48y 48x + 155 = 0

Giiy l mt cu kh hc, khng phi ai cng c th d dng gii n c.Th 3y= 9 x2 t (1) xung (2) ta c

y4 + 8xy2 12y2 16(9 x2) 48x + 155 = 0

y4 + 8xy2 + 16y2 12(y2 + 4x) + 11 = 0

y2 + 4x= 1y2 + 4x= 11

TH1 :

y2 + 4x= 11

9 x23

2

+ 4x= 11

x4

18x2 + 36x

18 = 0

x4 = 18(x 1)2

x2 32x + 32 = 0x2 + 3

2x 32 = 0



NguynM

inhTu

n


x=

3

2

18 1222

y= 12

2 6

36 24212

x=32

18 122

2 y=12

2 6

36 242

12

TH2 :y2 + 4x= 1

9 x2

3

2+ 4x= 1 x4 18x2 + 36x + 72 = 0

x2 6x + 12 x2 + 6x + 6 = 0 x= 3 3 y = 1 23Vy h c c thy 6 nghim nh trn

Mt thc mc nh l TH2 v sao x4 18x2 + 36x+ 72 = (x2 6x+ 12)(x2 + 6x+ 6). Tchnhn t kiu g hay vy ? Casio truy nhn t chng ? C th lm. Nhng thc ra phng trnhbc 4 c cch gii tng qut bng cng thc Ferrari. i vi v d trn ta lm nh sau

x4 18x2 + 36x + 72 = 0 x4 2ax2 + a2 = (18 2a) x2 36x + a2 72Ta phi tm asao cho v phi phn tch c thnh bnh phng. Nh th ngha l

182 = (18 2a) a2 72 a= 9Nh vy

x4 18x2 + 36x + 72 = 0 (x2 + 9)2 = 9(2x 1)2 (x2 6x + 12)(x2 + 6x + 6) = 0

Chi tit v gii phng trnh bc 4 cc bn c th tm d dng trn google. Gi ta tip tc cc

bi h. Tip theo l mt chm h s dng tnh n iu ca hm s kh d nhn.

Cu 108

x +

x2 + 1

y+

y2 + 1

= 1

y+ yx2 1 =

35

12

Gii

iu kin : x2 >1Khng th lm n c g t (2). T (1) ta nhn xt thy hai hm ging nhau nhng chngli dnh cht vi nhau, khng chu tch ri. Vy ta dt chng ra. Php lin hp s gip taPhng trnh (1) tng ng

x +

x2 + 1

y+

y2 + 1

y2 + 1 y

=

y2 + 1 y x +

x2 + 1 = y +

y2 + 1

Tch c ri nhng c v hai bn khng cn ging nhau na. Khoan !! Nu thay y2 = (y)2th sao nh. Qu tt. Nh vy c hai v u c dng f(t) =t+

t2 + 1v hm ny n iu

tng. T ta rt ra x=

y

Thay li vo (2) ta cy+

yy2 1 =

35

12



NguynM

inhTu

n

2.4 Cu 91 n cu 120 59

y thc ra l mt phng trnh kh kh chu. Thot tin khi thy loi ny ta s bnh phng2 v ln. iu kin bnh phng l y >0khi ta c

y2 + 2y2

y2 1+

y2

y2

1

= 35

122

y4 y2 + y2

y2

1

+ 2y2

y2 1=

35

122

n y kh r rng . t y2

y2 1 =t >0v phng trnh tng ng

t2 + 2t

35

12

2= 0

t=

49

12(L)

t=25

12

y2

y2 1 =25

12

y=

5

4

y= 53

i chiu iu kin bnh phng ch ly 2 gi tr dng.

Vy h cho c nghim : (x; y) =54;54 ,53;53

Cu 109

(4x2 + 1)x + (y 3)5 2y= 04x2 + y2 + 2

3 4x= 7

Gii

iu kin : y5

2 , x 3

4 Vit li phng trnh (1) nh sau

(4x2 + 1)x= (3 y)

5 2y (4x2 + 1)2x= (6 2y)

5 2y f(2x) =f

5 2y

Vi f(t) =t3 + tl hm n iu tng. T ta c 2x=

5 2y x 0thay vo (2) ta c

4x2 +

5

2 2x2

2+ 2

3 4x= 7

Gi cng vic ca ta l kho st hm s v tri trn 0;34v chng minh n n iu gim.

Xin nhng li bn cVi hm s v tri n iu gim ta c x=

1

2l nghim duy nht y = 2


1

2; 2

Hy k mi tng quan gia cc biu thc trong mt phng trnh va ta s t mc ch



NguynM

inhTu

n


Cu 110

y3 + y = x3 + 3x2 + 4x + 21 x2 y= 2 y 1

Gii

iu kin : 0 y 2,1 x 1Phng trnh (1) tng ng

y3 +y = (x + 1)3 + (x + 1) y = x + 1

Thay vo (2) ta c 1 x2 1 +x= 1 x 1

Phng trnh ny khng qu kh. t t =

1 +x+

1 x 1 x2 = t2 2

2 . Thay vo

phng trnh ta c

t2 22

=t 1

t= 0t= 2

1 x + 1 +x= 01 x + 1 +x= 2 x= 0, y = 1

Vy h cho c nghim :(x; y) = (0; 1)

Nhng bi ny thng s nng v gii phng trnh v t hn.

Cu 111x + 1 + x + 3 + x + 5 = y 1 + y 3 + y 5

x + y+x2 + y2 = 80

Gii

iu kin : x 1, y 5Phng trnh u c dng

f(x + 1) =f(y 5)Vi f(t) =

t +

t+ 2 +

t+ 4l hm n iu tng. T ta c y= x + 6thay vo (2) ta

c

x + x + 6 + x2 + (x + 6)2 = 80 x= 55 72 y= 55 + 5

2


5

5 72

;5

5 + 5

2

y ti a ra mt s cu h s dng tnh n iu ca hm s kh n gin. Ni l ngin v t mt phng trnh ta nhn thy ngay hoc mt cht bin i nhn ra dng cahm cn xt. Ti s cn gii thiu kh nhiu nhng bi cn bin i tinh t nhn ra dnghm, nhng cu sau ca cun sch.



NguynM

inhTu

n

2.4 Cu 91 n cu 120 61

Cu 112

x + 4

32 x y2 = 34

x +

32 x + 6y = 24

Gii

iu kin : 0 x 32C v y l mt h khc rc ri khi xut hin cn bc 4. Ta s dng cc nh gi giiquyt ci h nyCng 2 phng trnh cho nhau ta c

x +

32 x + 4x + 432 x= y2 6y+ 21

Hin nhin ta c : V P 12Gi ta tin hnh nh gi v tri. p dng bt ng thcCauchySchwarzcho v tri ta c

x +

32 x

(1 + 1)(x + 32 x) = 8

4x + 432 x (1 + 1)(x +32 x) 4VyV T V PDu bng xy ra khi (x; y) = (16; 3)

Ti cn mt cu tng ging bi ny nhng hi kh hn mt cht. Bn c c th gii n

Cu 113

2x + 2 4

6

x

y2 = 2

2

42x + 26 x + 22y = 8 + 2

Nghim : (x; y) = (2;

2)

Cu 114

x2(y+ 1)(x + y+ 1) = 3x2 4x + 1xy+x + 1 =x2

GiiBi ny c l khng cn suy ngh nhiu. C th y+ 1 ln (1) coi saoNhn thy x= 0khng l nghim. Phng trnh (2) tng ng

x(y+ 1) =x2 1 y+ 1 = x2 1

x

Thay ln (2) ta s c

x(x2 1)x +x2 1

x = 3x2 4x + 1

x= 2 y= 52

x= 1 y = 1Vy h cho c nghim : (x; y) = (1;1),

2;5

2



NguynM

inhTu

n


Cu 115

4xy+ 4(x2 +y2) + 3

(x + y)2 = 7

2x + 1

x + y = 3

Giiiu kin : x + y= 0y l mt bi h khng n gin cht no. Tuy nhin ta c mt nhn xt kh tt sau y :

a(x2 +y2) +bxy =k(x + y)2 + l(x y)2

Gi hy phn tch4x2 + 4y2 + 4xy= k(x + y)2 + l(x y)2Cn bng h s ta thu c : 4x2 + 4y2 + 4xy= 3(x + y)2 + (x y)2Nh vy tng s l t n ph tng-hiu chng ? Cng c c s khi2x= x + y + x y. Nhvy tng s b l th. Bin i h thnh

3(x + y)2 + (x y)2 + 3

(x + y)2 = 7

x + y+ 1

x +y+x y= 3

ng vi t ngay. mt cht 3(x+ y)2 + 3

(x + y)2 = 3

x + y+

1

x + y

2 6. Nh vy

cch t n ca ta s trit hn.t x + y+

1

x + y =a, x y = b ta thu c h mi

b2 + 3a2 = 13a + b= 3|a| 2

a= 2, b= 1

a= 12

, b=7

2(L)

x + y+

1

x + y = 2

x y = 1

x + y= 1x y= 1

x= 1y= 0


OK cha ? Tip tc thm mt cu tng t nh

Cu 116

x2 +y2 + 6xy 1(x y)2 +

9

8= 0

2y 1x y +

5

4= 0

Gii

iu kin : x =yH cho tng ng

2(x + y)2

(y x)2

1

(y x)2 +9

8= 0y x + 1

y x

+ (x + y) +5

4= 0

2(x +y)2 y x + 1y x2

+25

8 = 0y x + 1

y x

+ (x + y) +5

4= 0



NguynM

inhTu

n

2.4 Cu 91 n cu 120 63

t x + y= a, y x + 1y x =b, |b| 2ta c h mi

a + b=

5

42a2 b2 = 25

8

a=

5

4b= 5

2

y+ x=

5

4y

x=

2

y+ x= 5

4

y x= 12

x=13

8 , y=

3

8x=

7

8, y=

3

8


7

8;3

8

,

13

8;3

8

Ti s a thm 2 cu na cho bn c luyn tp

Cu 117

3(x2 + y2) + 2xy+ 1

(x y)2 = 20

2x + 1

x y = 5

Nghim : (x; y) = (2; 1),

410

3 ;

10 3

3

,

4 +

10

3 ;

3 103

Cu 118

(4x2 4xy+ 4y2 51)(x y)2 + 3 = 0(2x 7)(x y) + 1 = 0

Th ng no mt cht xem v sao li a c v ging 3 cu trn ?

Nghim :(x; y) = 5

3

2

;1 +

3

2 ,5 +

3

2

;1 3

2

Cu 119

2x

2 +x 1y

= 2

y y2x 2y2 = 2

Giiiu kin : y

= 0

Phng trnh (2) tng ng vi1

y x 2 = 2

y2



NguynM

inhTu

n


t a=1

yta chuyn h v

2x

2

+x a= 22a2 + a x= 2

x= 1, a= 1x= 1, a= 1

x=132

, a= 3 12

x=

3 1

2 , a=

1 32

Vy h cho c nghim : (x; y) = (1;1),13

2 ; 1 3

Cu 120

4x2 + y4 4xy3 = 14x2 + 2y2 4xy= 2

Gii

Hnh thc kh gn nh nhng cng rt kh chi. Mt cht tinh ta nhn thy y2 = 1 lnghim ca h. Thay vo v ta rt ra

P T(1) P T(2) y4 4xy3 2y2 + 4xy+ 1 = 0 (y2 1)(y2 4xy 1) = 0

Vi y= 1thay vo (2) ta tm c x= 0hoc x= 1Vi y= 1thay vo (2) ta tm c x= 0hoc x= 1Vi y2 = 4xy+ 1. Khng cn ngh nhiu, th tru b vo cho nhanh !!!

Ta rt ra x=y2 1

4y thay vo (2) ta c

4

y2 1

4y

2+ 2y2 + 1 y2 = 2 5y4 6y2 + 1 = 0

y= 1 x= 0y= 1 x= 0y= 1

5 x= 1

5

y= 1

5 x=

1

5

Vy h cho c nghim :(x; y) = (1; 1), (1;1), (0;1), (0;1),

15

; 15

,

1

5;

15



NguynM

inhTu

n

2.5 Cu 121 n cu 150 65

2.5 Cu 121 n cu 150

Cu 121

x4 + x3y+ 9y = y3x + x2y2 + 9xx(y3 x3) = 7

Gii

Khng cn bit T quc ni u, chin phng trnh u

P T(1) (x y)(x(x + y)2 9) = 0Vi x= y kt hp vi (2) r rng khng thaCn li ta kt hp thnh mt h mi

x (y3 x3) = 7x(x + y)2 = 9

y l mt bi ton kh quen thuc v hp dn tng xut hin trn bo THTT, cch lmph bin nht vn l "tru b"

Trc ht c nh gi x >0v rt ra y = 3

x3 +7

x. Thay xung ta c

x

x +

3

x3 +

7

x

2= 9 x3 + 2x 3

x6 + 7x2 + 3

x(x4 + 7)2 = 9

t v tri l f(x). Ta c

f(x) = 3x2 + 2

3

x6 + 7x2 + 6x6 + 14x2

3 3

(x6 + 7x2)2

+

1

3.9x8 + 70x4 + 49

3

x2(x4 + 7)4>0

Vyf(x) = 9c nghim duy nht x= 1 y= 2Vy h cho c nghim : (x; y) = (1; 2)

Tip theo ti xin gii thiu cho cc bn mt s cu h s dng Bt ng thc Minkowskigii. Bt ng thc Minkowski l mt bt ng thc khng kh v cng thng c dng,bt ng thc cp n vn di ca vect trong khng gian m sau ny hc sinh quen

gi n l bt ng thc V ectorVi hai vectu ,v bt k ta lun c

|u |+ |v | |u +v |Nu ta ha 2 vecto ny ta s thu c

a12 +b12 +

a22 + b22

(a1+ a2)2 + (b1+ b2)

2

ng thc xy ra khi (a1, a2)v(b1, b2)l 2 b t ly l mt h qu hay dng trong gii h

Th khi no nhn vo mt bi h ta c th ngh n s dng Bt ng thc Minkowski. Thngkhi nhn thy tng hai cn thc m bc ca biu thc trong cn khng vt qu 2 th ta cth chn hng ny. Ti s nu 3 v d bn c hiu r hn



NguynM

inhTu

n


Cu 122

3x + 4y= 26x2 + y2 4x + 2y+ 5 +

x2 + y2 20x 10y+ 125 = 10

Gii

tng s dng hin r ri. Bc u tin ta lm l phn tch biu thc trong cnthnh tng cc bnh phng . V tri ca (2) khi s l

(x 2)2 + (y+ 1)2 +

(x 10)2 + (y 5)2

Tuy nhin nu ta s dng Bt ng thc Minkowskingay by gi th n s l

V T

(x 2 +x 10)2 + (y+ 1 +y 5)2

Khng phi 10 na m l mt biu thc kh phc tp. Khi ta phi xem li cch vit cc

bnh phng ca mnh nu l hng s v phi th khi cng vo ta phi lm trit tiu n i. Vy cn phi vitnh sau

V T =

(x 2)2 + (y+ 1)2+

(10 x)2 + (5 y)2

(x + 2 + 10 x)2 + (y+ 1 + 5 y)2 = 10

Ok ri. ng thc xy ra khi 10 x

x 2 =5 yy+ 1

3x 4y = 10Kt hp (1) d dng gii ra (x; y) = (6; 2)

Nh ta thy, s dng khng kh. Tuy nhin ci kh y chnh l ngh thut i du vsp xp cc hng t ca bnh phng ta t c mc ich

Cu 123

x2 2y2 7xy= 6x2 + 2x + 5 +

y2 2y+ 2 =

x2 + y2 + 2xy+ 9

Gii

Xt phng trnh (2) ta c

V T =

(x + 1)2 + 22 +

(y 1)2 + 12

(x + y)2 + 32 =V P

ng thc xy ra khi x + 1 = 2(y 1) x= 2y 3Thay vo (1) v ta d dng gii ra (x; y) =

5

2;1

4

, (1;1)



NguynM

inhTu

n

2.5 Cu 121 n cu 150 67

Cu 124

2x2 + 6xy+ 5y2 + 5 =

2x2 + 6xy+ 5y2 + 14x + 20y+ 25x4 + 25y2 2 = 0

Gii

By gi nu chuyn cn sang v tri, hng s sang v phi l cht d. Mu cht y l g ?S 5 chng ? ng vy, ta phn tch 5 =

32 + 42 s dng bt ng thc Minkowski. Tuy

nhin cc i du v sp xp s hng nh th no. Ci ta phi quan tm n v phi chn la cho ph hp. y s l

V T =

(x + y)2 + (x + 2y)2 +

42 + 32

(x + y+ 4)2 + (x + 2y+ 3)2 =V P

ng thc xy ra khi x + y

4 =

x + 2y

3 x= 5y

Thay vo (2) v ta d dng gii ra (x; y) =

1;15

,1;15

Cu 125

2y(x2 y2) = 3xx(x2 +y2) = 10y

Gii

Mt h a v dng ng cp r rng. Tuy nhin, ta hy x l s b h ny loi mt strng hpT (2) d thy x.yphi cng du, m nu th (1) x2 y2Trc ht x= y = 0l mt nghim ca hNhn cho 2 phng trnh cho nhau ta c

20y2(x2 y2) = 3x2(x2 +y2) (x 2y)(2y+ x)(5y2 3x2) = 0

V x v y cng du nn nn t y ta suy ra x= 2yhoc x=

5

3y

n y ch vic thay vo (1). Xin nhng li cho bn cVy h cho c nghim :(x; y) = (0; 0), (2; 1), (2,1),

4

30375

6 ;

4

135

2

,

4

30375

6 ;

4

135

2



NguynM

inhTu

n


Cu 126

7 +x +

11 y = 67 +y+

11 x= 6

Gii

Cng 2 phng trnh cho nhau ta c

7 +x +

11 x +

7 +y+

11 y= 12

p dng bt ng thc Cauchy Schwarzcho v tri ta c

V T

(1 + 1)(7 + x + 11 x) +

(1 + 1)(7 +y+ 11 y) = 12Du bng xy ra khi (x; y) = (2; 2)

Cu 127

2x2y2 + x2 + 2x= 22x2y x2y2 + 2xy= 1

Gii

Bin i 1 t, h cho tng ng

2x2y2 + (x + 1)2 = 3

2xy(x + 1) x2

y2

= 1 (xy+ x + 1)2 = 4

xy= 1 xxy= 3 x

Vi xy= 1 xthay vo (1) ta c

2(1 x)2 + x2 + 2x= 2

x= 0(L)

x=2

3 y=1

2

Vi xy= 3 xthay vo (2) ta c

2(x + 3)2

+ x2

+ 2x= 3 x=

8

3y =

1

8x= 2 y = 12


2

3;1

2

,

8

3;1

8

,

2;1

2



NguynM

inhTu

n

2.5 Cu 121 n cu 150 69

Cu 128

(x 1)(y 1)(x + y 2) = 6x2 + y2 2x 2y 3 = 0

Gii

Bi ny tng t n ph r rngt x 1 =a, y 1 =b ta a v h sau

ab(a +b) = 6a2 +b2

tuyển chọn 410 hệ phương trình và các phương pháp giải hệ phương trình

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