tuyen tap 410 cau he phuong trinh

Nguyn Minh TunSinh vin K62CLC - Khoa Ton Tin HSPHN

TUYN CHN 410 BI H

PHNG TRNH I S

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

NG

H Ni, ngy 9 thng 10 nm 2013

Mc 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 . . . . . . . . . . . . . . . . . . 5

1.2 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

Li 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, cuh 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

Chng 1

Mt s phng php v cc loi h cbn

1.1 Cc phng php chnh gii h phng trnh

I. 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

6 Mt s phng php v cc loi h c bn

1.2 Mt s loi h c bn

A. H phng trnh bc nht 2 n

I. Dng

{ax+ by = c (a2 + b2 6= 0)ax+ by = c (a2 + b2 6= 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 x v y cho 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 III. Du hiui vai tr ca x v y cho 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 3

{ax3 + 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

Chng 2

Tuyn 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 lnhri. 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 + 2yx2 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

8 Tuyn tp nhng bi h c sc

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

6y2 = 6[y = 1, x = 3y = 1, x = 3

TH3 : x = 4y thay 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

),

(4

6

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.PT (1) PT (2) y2 + 4y = 0

y = 0 x = 3

7

2

y = 4 x = 3

7

2

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

(37

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 v xy = b ta c h mi


NguynMinhTun

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

Gii

H 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 = a v xy = b ta 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

Gii

y 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 = a v 2x+ 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

Gii

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


NguynMinhTun


(2) x+ y + 2 + 2xy + x+ y + 1 = 16

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

(2) 3 +xy + 2 + 2xy +

xy + 4 = 16 xy = 3

{xy = 9x+ y = 6

x = y = 3

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

Cu 8

{ x+ 5 +

y 2 = 7

x 2 +y + 5 = 7

Gii

i 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 = yThay 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 = 16

x+ y + 2xy = 16

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

Thay li ta c : 2x = 4 x = 4


NguynMinhTun

2.1 Cu 1 n cu 30 11


Cu 10

{6x2 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) x p 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 =

35

Vy h cho c nghim (x; y) =

(1

3;2

2

3

), (0, 1),

(4

5;3

5

)

Cu 11

{x 2y xy = 0x 1 +4y 1 = 2

Gii

Phng trnh u l dng ng cp ri

iu kin x 1, y 14

T phng trnh u ta c :(

x+y) (

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

x 1 +x 1 = 2 x = 2Vy 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 0Phng trnh u tng ng

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


NguynMinhTun


Vi x = y loi v theo iu kin th x, y phi 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 = 5Vy h cho c nghim (x; y) = (5; 2)

Cu 13

{ x+ 1 +

y + 2 = 6

x+ y = 17

Gii

iu kin x, y 1H cho tng ng

{ x+ 1 +

y + 2 = 6

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

tx+ 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 = 0 th suy ra : (5x+ 4) (4 x) = 0[x = 4

x = 45

Vi y = 2x+ 4 th 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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NguynMinhTun

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 = 0y = 12

(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

PT (1) PT (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 =941

20, x = 1 +

41

10

y =9 +

41

20, x =

41 110

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

(1 +

41

10;941

20

),

(41 110

;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 y2

x+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) (xy + 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 = 1000 ta 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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NguynMinhTun

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 13

)(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 x v y nh nhauCho y = 1000 ta c C = x3 7x2 2989992x 1983039984Phn tch C= (x 1999) (x+ 996)2Cho 1999 = 2y 1 v 996 = y 4C = (x 2y + 1) (x+ y 4)2

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

V d 5 : E = x3y + 2x2y2 + 6x3 + 11x2y xy2 6x2 7xy y2 6x 5y + 6Bc ca y nh hnCho 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 = 1000 ta 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 + 3x 5725

= y(3x+ 1)

Gii


NguynMinhTun


Li gii gn p nht ca bi trn l

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

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

(2

5;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.PT (1) 15.PT (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:

{a1x

2 + b1y2 + c1xy + d1x+ e1y + f1 = 0

a2x2 + b2y

2 + 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 PT (1) + k.PT (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 k l nghim ca phng trnh sau vi a 6= 0

cde+ 4abf = ae2 + bd2 + fc2

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 + 56kS k s 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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NguynMinhTun

2.1 Cu 1 n cu 30 17

Nh vy l PT (1) 1549.PT (2) hay 49.PT (1) 15.PT (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 nhiungun.

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 = 33

5.

{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 = 0x2 + 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

PT (1) 3.PT (2) (x 2)3 = (y + 3)3 x = y + 5Thay 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


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- Bc ca x v y l nh nhau- Cc bin x,y c 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) PT (1) + a.PT (2) ac v dng hng ng thc A3 = B3

PT (1) + a.PT (2) x3 + 2ax2 4ax y3 + 3ay2 + 9ay 35 = 0Cn tm a sao cho v tri c dng (x+ )3 (y + )3 = 0

Cn bng ta c :

3 3 = 353 = 2a32 = 4a

a = 3 = 2 = 3

Vy PT (1) 3.PT (2) (x 2)3 = (y + 3)3OK ?? Th mt v d tng t nh

Gii h:

{x3 + y3 = 914x2 + 3y2 = 16x+ 9y

Gi : PT (1) 3.PT (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.PT (2) PT (1) s c nhn t x 1

C th l (x 1) (y2 (x+ 3) y + x2 x 2) = 0TH1 :x = 1 thay vo th v nghimTH2: 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


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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 h

theo n y v hi vng n s li ng vi x no . Tht vy, l x = 12. Tip tc thay n

vo h v ta s rt ra :

2PT (2) PT (1) (2x+ 1) (y2 (x 1) y + x2 x+ 2)TH1 : 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 23Mt 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 PT (1) 21PT (2) (x 2)

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

TH1 : x = 2 y = 1TH2 :

{2y2 + 2xy + 4y 17x 126 = 0x2 + y2 + 2x 2y 7 = 0

H ny c cch gii ri nh ??3.PT (2) PT (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.


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Cu 25

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

Gii

Li gii ngn gn nht ca bi trn l :

PT (1) + 3.PT (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 = 1 phng trnh 2 s khng cnphn y v c v 2 phng trnh s tng ng. Khi thay x = 1 h cho tr thnh{ 3y2 + 48 = 0

y2 16 = 0

Hai phng trnh ny tng ng. Tri thng ri !! Vy x = 1 chnh l 1 nghim cah v t h th hai ta suy ra ngay phi lm l PT (1) + 3.PT (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 : PT (1) + 3.PT (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 : PT (1) + 3.PT (2) (2y + 5)(

3

(x+

1

2

)2+

(y +

5

2

)2)= 0


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2.1 Cu 1 n cu 30 21

Cu 28

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

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

Cu 29

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

Gi : PT (1) + 3.PT (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 dngbng 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; 0 v (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


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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) .PT (1) + 9.PT (2)Nh vy

20 (y 1) .PT (1) + 9.PT (2) (x+ 2y) (18x2 + 15xy 60x 10y2 80y) = 0TH1 : x = 2y thay 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),(

151452

; 11145),

(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 a vo 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,y

2,y, ....}

Tt nhin khng th c nghim x = 12hay x = 1 c. 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).PT (1) 3(y 6).PT (2). R rng l qu phc tp. Loi ci ny.* Vi x = y thay vo h ta c

{y2 = 03y3 = 0

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

(x+ y)(2x2 + 6xy (3y2 + 27y + 20)) = 0

Cch 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 kthp 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.


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NguynMinhTun

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

PT (1) (y 1).PT (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 = 3 l 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 bv 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 lm

ng. 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 = y thay 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 = 1 thay vo (2) ta c

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


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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 32y 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 3t vi 1 t 1 th f (t) = 3t2 3 0Suy ra f(t) n iu v t suy ra x = y 1 thay 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) 134

Ta li c : g(y) = 3y(2 y) 3y + 2 y

2= 3

Vy f(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) = t33t li 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 = 16

2

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

(16

2;

16

2

),

( 1

6

2; 1

6

2

)


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Cu 37

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

Gii

Nhn thy x = 0 khng l nghim. H cho tng ng3y + 1 =

1

x33

x+ 2 = y3

y = 1x

Thay li (1) ta c

2x3 + 3x2 1 = 0[x = 1 y = 1x =

1

2 y = 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 = 0 l nghim ca h. Thay li v ta s c

2PT (1) + PT (2) y(x+ y + 5)(x+ y 3) = 0 y = 0x = 5 yx = 3 y

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

(y + 5)2 + y2 y2 5y + 1 = 4y V L

Tng t vi x = 3 y cng v nghimVy h cho v nghim


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2.2 Cu 31 n cu 60 27

Cu 39

{ x+ y x y = y

2x2 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 2x2 y2 = y

2

4

n y nhn thyx2 y2 theo (2) bng 3. Vy suy ra

2x 6 = y2

4 y2 = 8x 24

Thay vo (2) ta c

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

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

Cu 40

xy + 1 = 5

2

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

a2 1 b = 52

b2 1 + 2a(a2 4) = 34

Ta th b =7

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

(7

2 a2

)2+ 2a(a2 4) 1

4= 0

a = 3 b = 112

(L)

a = 2 b = 12

(L)

a = 1 b = 52

(TM)

a = 2 b = 12

(L)

{x = 0

y = 34


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(0;3

4

)

Cu 41

{(x2 + xy + y2)

x2 + y2 = 185

(x2 xy + y2)x2 + y2 = 65Gii

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 + y2

Cng 2 phng trnh cho nhau ta c

2(x2 + y2)x2 + y2 = 250

x2 + y2 = 5

Khi 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

xxy + 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)


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NguynMinhTun

2.2 Cu 31 n cu 60 29

Cu 43

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

Gii

Dng UCTPT (1) 3.PT (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 = 0 khng 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 95

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 +

5) x = 1

4

(3 +

5)


(1

4

(35) ;3

2

(5 3)) ,(1

4

(3 +

5)

;3

2

(3 +

5))


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Cu 45

(x+ y)

(1 +

1

xy

)= 5

(x2 + y2)

(1 +

1

x2y2

)= 9

Gii

iu kin : xy 6= 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+1

x= 2, y +

1

y= 3

x+1

x= 3, y +

1

y= 2

x = 1, y = 3

5

2

x =35

2, y = 1


(1;

352

),

(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 = 6a = 6, b = 12

{x2 + x = 6y2 + y = 12{x2 + 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


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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 = 2y =

x2 165x

Trng hp 2 thay vo (2) s l

(x2 16)225x2

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

xy2 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 = 5Thay 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)


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Cu 49

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

Gii

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

PT (1) + 2.PT (2) (x2 + y)2 2(x2 + y) 35 = 0

TH1 : x2 + y = 7 x2 = 7 y thay (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 = 5

4

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

Gii

y 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 = 54

(x2 + y)2 + xy = 54

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

a+ b+ ab = 5

4

a2 + b = 54

a = 0, b = 54a = 1

2, b = 3

2

{x2 + y = 0

xy = 54

x2 + y = 12

xy = 32

x = 3

5

4, y = 3

25

16

x = 1, y = 32


(3

5

4; 3

25

16

),

(1;3

2

)


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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 + 1 v x + y. Tuy nhin y chen vo khin h tr nnkh chu. Hy dit y i . Cch tt nht l chia khi m y = 0 khng 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. tx2 + 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 = 0 khng 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 2y

x= 5

ty

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

)


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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 = 0TH2 : 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 = 0 khng l nghim ca h. PT(1) khi s l

2(yx

)+(yx

)3= 2x+ x3 y

x= 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(TM)

x = 3, y = 3(TM)

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

Ta s n mt cu tng t n


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2.2 Cu 31 n cu 60 35

Cu 55

{x5 + xy4 = y10 + y6

4x+ 5 +y2 + 8 = 6

Gii

iu kin : x 54

Thy y = 0 khng 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)cho y2 sau khi kim tra y = 0 khng 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 xy

= 13{a+ b = 7a2 b = 13

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

x+1

y= 4

x = 3y x+1

y= 5

x = 12y

[x = 1, y =

1

3x = 3, y = 1


(1;

1

3

), (3; 1)

Tip tc ta n thm mt cu tuyn sinh na


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Cu 57

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

Gii

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

4 + 2x2(xy) + x2y2 = 2x+ 9

xy =6x+ 6 x2

2

T (1) s tng ng

x4 + x2(6x+ 6 x2) +(

6x+ 6 x22

)2= 2x+ 9

[x = 4x = 0

[y =

17

4V L


(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 = 2b 0

a = 1, b = 1a = 13, b = 3 +3a =

3 1, b = 33 y = 0y = 63 11y = 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), (6

3 11; 63 11), (63 11;63 11)


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2.2 Cu 31 n cu 60 37

Cu 59

{ xy +

1 y = y

2yx 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

2yx 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 + y19 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 2x v 17 4x2, y19 9y2 lin quan n 3y v 19 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 19

6= 3

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

TH1 :

{2x+

17 4x2 = 5

3y +

19 9y2 = 5

[x =

1

2x = 2

y =513

6

TH2 :

{2x+

17 4x2 = 3

3y +

19 9y2 = 7 (Loi)


(1

2;5 +

13

6

)(1

2;513

6

)(2;

5 +

13

6

)(2;

5136

)

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


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2.3 Cu 61 n cu 90

Cu 61

{x

5 x2 + y5 4y2 = 15 x2 +5 4y2 = 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 1 v y3 1 t (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 =1

3

25, x =

33

25

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

(1

3

25;

33

25

)

Cu 63

x+

x2 y2

xx2 y2 + xx2 y2

x+x2 y2 =

17

4

x(x+ y) +x2 + xy + 4 = 52

Gii

iu kin : x 6= 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 y2

)2x2 (x2 y2) +

(xx2 y2)2x2 (x2 y2) =

17

4 2 (2x

2 y2)y2

=17

4 y = 4x

5


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2.3 Cu 61 n cu 90 39

Tip tc khai thc (2). D thy tx2 + xy + 4 = t 0 th (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 64

{ x+y +

xy = 2

y +x

y x = 1

Gii

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+ 2x2 y = 4

x2 y = 2 x x2 y = x2 4x 4 4x y = 4

Lm tng t phng trnh (2) ta s c : 4x 4y = 1. Kt hp 2 kt qu li d dng tmc x,y


(17

12;5

3

)

Cu 65

x+

2xy3x2 2x+ 9 = x

2 + y

y +2xy

3y2 2y + 9 = y

2 + x

Gii

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

x2 + y2 = 2xy

(1

3x2 2x+ 9 +

13y2 2y + 9

)

T ta nhn xt c nghim th xy 0 v l 3t2 2t+ 9 2 nn ta nh gi

x2 + y2 2xy(

1

2+

1

2

) (x y)2 0


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

Cu 66

{6x

y 2 = 3x y + 3y

2

3x+

3x y = 6x+ 3y 4

Gii

iu kin : y 6= 0 , 3x y, 3x+3x y 0Phng trnh (1) khi s tng ng

6x 2y = y

3x y + 3y2 2 (3x y) y

3x y 3y2 = 0[

3x y = y

3x y = 3y2

TH1 :

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

2y2 + y y = 2 (y2 + y)+ 3y 4 { 2y2 + 7y 4 = 0

y 0 y = 4 x = 4

TH2 :

3x y = 3y2. T y suy ra y 0 v 3x = 9y

2

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

c y =8

9 x = 8

9

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

8

9;8

9

)

Cu 67

{(3 x)2 x 2y2y 1 = 03x+ 2 + 2

y + 2 = 5

Gii

iu kin : x 2, y 12

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 + x n iu tng. T suy ra

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

3

5 2y + 2y + 2 = 5

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

a = 1, b = 2

a =365

4, b =

23 +

65

8

a =

65 3

4, b =

23658


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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 68

{ 1 + 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+ y thay vo (1) ta c

2

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

1 + xy +xy = 2(V L)


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

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

x2 + y2= 0


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t z = x+ yi khi 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 11yx2 + 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.


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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 = 1 l nghim ca h . Ta suy ra

PT (1) + PT (2) (x 1)(4y2(x+ 1) + 6xy 9) = 0TH1 : x = 1 y = 3TH2 : 4y2(x+ 1) + 6xy 9 = 0V x = 0 khng l nghim. Suy ra 4y2x(x+ 1) + 6x2y 9x = 0 (*)V sao nhn x vo 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 = 0TH1 : y = 0 v nghim

TH2 : x = 12 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

23xy = x+ y + 1

Gii

iu kin x, y 6= 1Bi ton ny c kh nhiu cch gii. Ti xin gii thiu cch p nht ca bi ny

p dng Bt ng thc AM GM cho 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),

(1

3;1

3

)

Cu 76

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

x2 + 2y + 1

y(y x) = 3 3y

Gii

iu kin : x2 + 2y + 1 0Khng lm n g c t (2). Th bin i (1) xem sao. PT(1) tng ng

4y2 4yx2 + 2y + 1 + x2 + 2y + 1 = x2 2xy + y2

(2y

x2 + 2y + 1

)2= (x y)2


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[

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 nyBy gi x l hai trng hp kia th no ? Chc bnh phng thi. Tt qu ! Phng trnh sch cn li xy v y m 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(TM)

x =415

51, y =

17

3(TM)

TH2 :x2 + 2y + 1 = x+ y

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

x+ y 02xy = y2 + 2y + 1xy = y2 + 3y 3

[x = 1, y = 1

x =41

21, y = 7

3(L)


(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 nythng 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

Gii

Hnh 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 y 73

(**) 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 P nn 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 x2

th s c lp c x v y v hi vng s ra c iu g.Nhn thy x = 0 khng l nghim. Chia 2 v ca (2) cho x2 ta c

2y + 2y

4y2 + 1 =1

x+

1

x

1

x2+ 1


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R rng 2 v u c dng f(t) = t + tt2 + 1 v hm ny n iu tng. Vy t ta suy ra

c 2y =1

xthay vo (1) ta c

x3(

1

x2+ 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 = 1 l nghim duy nht


(1;

1

2

)

Cu 87

{ 7x+ y +

2x+ y = 5

2x+ 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 = b xem

Nhng cn x y th 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 = 8

5

Vy h cho tng nga+ b = 5

b+3a2

5 8b

2

5= 2

a, b 0

a =

15772

b =

77 5

2

7x+ y =

151 15772

2x+ y =51 577

2

x = 10

77

y =1177

2


(1077; 11

77

2

)

Mt cch khc cng kh tt. t

7x+ y = a,

2x+ y = b v 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


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2.3 Cu 61 n cu 90 49

Cu 88

{ 11x y y x = 1

7y 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=

13x 2

27y

1 =13x

+2

27y

1x+ y

=

(13x 2

27y

)(13x

+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

PT (1) + i.PT (2) (a+ bi) + a bia2 + b2

=23

+4

27i

t z = a+ bi phng trnh cho tr thnh

z +1

z=

23

+4

27i z a, b x, y


NguynMinhTun



(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

4x

(1

4+

2x+y

x+ y

)= 2

4y

(1

4 2x+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


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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 = sinb vi a, b

[pi

2;pi

2

]Phng trnh u tng ng

sinacosb+ sinbcosa = 1 sin(a+ b) = 1 a+ b = pi2


(1 sina)(1 + sinb) = 2 (1 sina)(1 + cosa) = 2[a = pi

2a = 0

[b = pi

b =pi

2

[x = 1, y = 0(L)x = 0, y = 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 =2y

1 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 ra

t x = tan vi (pi

2;pi

2

). T PT(2) ta s c

y =3 tan tan3

1 3tan2 = tan 3

M nh th theo (1) ta s c

x =2 tan 3

1 tan23 = tan 6


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T suy ra

tan = tan 6 = kpi5 =

{2pi

5;pi

5; 0;

pi

5;2pi

5

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

(tan2pi

5; tan

6pi5

),

(tanpi5

; tan3pi

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+ y v 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 = 5x+ y = 1

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

x = 5

21

2, y =

5 +

21

2

x =5 +

21

2, y =

5212


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2.4 Cu 91 n cu 120 53

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

(521

2;521

2

)

Tc gi ca n rt kho lo trn nhiu ln cch t n tng tch vo mt h, gy nhiu khkhn 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 Schwarz cho 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 210 tc l phi chng minhsin2x+

1

sin2x+

cos2x+

1

cos2x

10

V T =

(sinx 1

sinx

)2+(

2)2

+

(cosx 1

cosx

)2+(

2)2

(

1

sinx+

1

cosx (sinx+ cosx)

)2+(

2

2)2

Hin nhin ta c sinx+ cosx 2 nn1

sinx+

1

cosx (sinx+ cosx) 4

sinx+ cosx

2 42

2 =

2

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

pi

4+ k2pi


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Cu 99

{xxx = yy + 8y

x 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=

2x

y+ 2

y(

x2 + 1 1) = 3x2 + 3Gii

iu kin : x > 0, y 6= 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 = 2x thay vo (2) ta c

2x(

x2 + 1 1)

=

3x2 + 3(

2x

3)

x2 + 1 = 2xx2 + 1 =

2x

2x3R 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)


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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 = 2 l 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 cy = 0, x = 0

y = 10

10

3, x = 20

10

3

y = 10

10

3, x = 20

10

3


(20

10

3;10

10

3

)


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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 = 27a = 9, b =

1

3

{x2 + y2 = 82x = 7y{x2 + y2 = 10x = 3y

y = 4 253 , x = 14

2

53x = 3, y = 1

Vy h cho c nghim : (x; y) = (3;1)(14

2

53;4

2

53

)

Cu 104

x+

1

y+x+ y 3 = 3

2x+ y +1

y= 8

Gii

iu kin : y 6= 0, x+ 1y 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 = 2a = 2, b = 1

x+1

y= 1

x+ y 3 = 4 x+1

y= 4

x+ y 3 = 1

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

10, y = 310

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

Vy h cho c nghim : (x; y) = (3; 1), (5;1)(410; 310)


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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 = 3 l nghim (ci ny gi li nh, ti khng gii thch na), thay y = 3 vo hta c {

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

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

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

Gii

y 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


NguynMinhTun


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 x23

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

(x2 6x+ 12) (x2 + 6x+ 6) = 0 x = 33 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 72

Ta phi tm a sao 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 ccbi 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 +y

x2 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 + 1 v hm ny n iu

tng. T ta rt ra x = yThay li vo (2) ta c

y +yy2 1 =

35

12


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NguynMinhTun

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 > 0 khi ta c

y2 +2y2y2 1 +

y2

y2 1 =(

35

12

)2 y

4 y2 + y2y2 1 +

2y2y2 1 =

(35

12

)2

n y kh r rng . ty2y2 1 = t > 0 v phng trnh tng ng

t2 + 2t(

35

12

)2= 0

t = 4912(L)t =

25

12

y2

y2 1 =25

12

y = 54y = 5

3

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


(5

4;5

4

),

(5

3;5

3

)

Cu 109

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

3 4x = 7

Gii

iu kin : y 52, x 3

4Vit 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 + t l hm n iu tng. T ta c 2x =

5 2y x 0 thay 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;

3

4

)v chng minh n n iu gim.

Xin nhng li bn c

Vi 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


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Cu 110

{y3 + y = x3 + 3x2 + 4x+ 2

1 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 111

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

c

x+ x+ 6 + x2 + (x+ 6)2 = 80 x = 5

5 72

y = 5

5 + 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.


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2.4 Cu 91 n cu 120 61

Cu 112

{ x+ 4

32 x y2 = 34x+

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 + 21Hin nhin ta c : V P 12Gi ta tin hnh nh gi v tri. p dng bt ng thc CauchySchwarz cho v tri ta c

x+

32 x

(1 + 1)(x+ 32 x) = 84x+ 4

32 x

(1 + 1)(x+

32 x) 4Vy V 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 = 22

4

2x+ 2

6 x+ 22y = 8 +2

Nghim : (x; y) = (2;

2)

Cu 114

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

Gii

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

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

Thay ln (2) ta s c

x(x2 1)(x+

x2 1x

)= 3x2 4x+ 1

[x = 2 y = 5

2x = 1 y = 1

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

2

)


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Cu 115

4xy + 4(x2 + y2) +

3

(x+ y)2= 7

2x+1

x+ y= 3

Gii

iu kin : x+ y 6= 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 tch 4x2 + 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 khi 2x = 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 6= yH cho tng ng

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

9

8= 0(

y x+ 1y x

)+ (x+ y) +

5

4= 0

2(x+ y)2

(y x+ 1

y x)2

+25

8= 0(

y x+ 1y x

)+ (x+ y) +

5

4= 0


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NguynMinhTun

2.4 Cu 91 n cu 120 63

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

a+ b = 5

4

2a2 b2 = 258

a =

5

4

b = 52

{y + x =

5

4y x = 2y + x =

5

4

y x = 12

x = 138 , y = 38x =

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;310

3

)

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) =

(53

2;1 +

3

2

),

(5 +

3

2;13

2

)

Cu 119

2x2 + x1

y= 2

y y2x 2y2 = 2

Gii

iu kin : y 6= 0Phng trnh (2) tng ng vi

1

y x 2 = 2

y2


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t a =1

yta chuyn h v

{2x2 + x a = 22a2 + a x = 2

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

x =13

2, a =

3 12

x =

3 12

, a =13

2

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

2; 13

)

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

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

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

Ta rt ra x =y2 1

4ythay 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 =15 x = 1

5

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

15

; 15

),

( 1

5;

15

)


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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

PT (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 > 0 v rt ra y = 3x3 +

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

(3x6 + 7x2 +

6x6 + 14x2

3 3

(x6 + 7x2)2

)+

1

3.9x8 + 70x4 + 49

3x2(x4 + 7)4

> 0

Vy f(x) = 9 c 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 Minkowski gii. 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 quengi n l bt ng thc V ectorVi hai vect u ,v bt k ta lun c

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

a12 + b12 +

a22 + b2

2

(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 hTh khi no nhn vo mt bi h ta c th ngh n s dng Bt ng thcMinkowski. 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


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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 Minkowski ngay 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 ccbnh 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 khi10 xx 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)


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NguynMinhTun

2.5 Cu 121 n cu 150 67

Cu 124

{ 2x2 + 6xy + 5y2 + 5 =

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

x4 + 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 khix+ y

4=x+ 2y

3 x = 5y

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

(1;1

5

),

(1; 1

5

)

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.y phi cng du, m nu th (1) x2 y2Trc ht x = y = 0 l 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 = 2y hoc 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

)


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Cu 126

{ 7 + x+

11 y = 6

7 + 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 Schwarz cho 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 = 32xy(x+ 1) x2y2 = 1 (xy + x+ 1)

2 = 4[xy = 1 xxy = 3 x

Vi xy = 1 x thay vo (1) ta c

2(1 x)2 + x2 + 2x = 2[x = 0(L)

x =2

3 y = 1

2

Vi xy = 3 x thay vo (2) ta c

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

x = 83 y = 18x = 2 y = 1

2


(2

3;1

2

),

(8

3;1

8

),

(2; 1

2

)


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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 = 5

[a = 1, b = 2a = 2, b = 1

[x = 2, y = 3x = 3, y = 2

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

Cu 129

{(x y)(x2 + xy + y2 + 3) = 3(x2 + y2) + 24x+ 2 +

16 3y = x2 + 8

Gii

iu kin : x 2, y 163

Phng trnh u tng ng

x3 y3 + 3(x y) = 3(x2 + y2) + 2 (x 1)3 = (y + 1)3 y = x 2

Thay vo (2) ta c4x+ 2 +

22 3x = x2 + 8

y l mt phng trnh v t khng hn l d xi. Ci hay ca bi ny yPhng trnh cho tng ng

4

(x+ 2 x+ 4

3

)+

(22 3x 14 x

3

)= x2 x 2

4

9(x+ 2) x2 8x 169

(x+ 2 +

x+ 4

2

)+

9(22 3x) x2 + 28x 1969

(22 3x+ 14 x

3

) = x2 x 2

(x2 x 2)1 + 4

9

(x+ 2 +

x+ 4

3

) + 19

(22 3x+ 14 x

3

) = 0 [ x = 1x = 2

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

Cu hi t ra l v sao li chn c nhng biu thc kia lin hp. Mt s tnh c chng?Khng ! L c mt phng php ! Ti xin vit 1 bi nh y tip


NguynMinhTun


i mt phng trnh v t, phng php hay dng nht l nhn lng lin hp. Tuynhin, nhn lin hp cng cn mt cht k thut. i vi bi ny ta tin hnh nh sauNhm hoc Casio ta thy phng trnh c nghim x = 1;x = 2i vi phng trnh c 2 nghim tr nn th cch thm bt hng s vo mi cn ri lin hpl khng ph hp, y ta khng thm bt hng s m thm hn mt biu thc ax+ b no Trc ht vi

x+ 2 nh

Vi x = 1 thay vo cn c gi tr bng 1, thay vo biu thc thm ta c a+ b = 1Vi x = 2 thay vo cn c gi tr bng 2, thay vo biu thc thm ta c 2a+ b = 2

Gii h ny ra ta c a =1

3, b =

4

3

Vy biu thc cn chn vo lx+ 4

3Tng t vi

22 3x . OK ???

Cu 130

{ x2 + x+ y + 1 + x+

y2 + x+ y + 1 + y = 18

x2 + x+ y + 1 x+y2 + x+ y + 1 y = 2Gii

tng t n cng l rit

x2 + x+ y + 1 +

y2 + x+ y + 1 = a 0, x+ y = b ta c h mi{

a+ b = 18a b = 2

{a = 10b = 8

{x+ y = 8x2 + 9 +

y2 + 9 = 10

y l mt h kh n gin v c nhiu cch. Ti u nht lx2 + 32 +

y2 + 32

(x+ y)2 + (3 + 3)2 = 10

ng thc xy ra khi (x; y) = (4; 4)

Cu 131

{12x+ 3y 4xy = 16

4x+ 5 +y + 5 = 6

Gii

iu kin : x 54, y 5, xy 0

T phng trnh u ta thy ngay x, y > 0Phng trnh u tng ng

12x+ 3y = 16 + 2

4xy 16 + (4x+ y) 4x+ y 8T phng trnh (2) ta li c

4x+ 5 +

y + 5

(1 + 1)(4x+ y + 10) 6

ng thc xy ra khi {4x = y4x+ y = 8

{x = 1y = 4



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NguynMinhTun

2.5 Cu 121 n cu 150 71

Cu 132

{2x+ (3 2xy)y2 = 32x2 x3y = 2x2y2 7xy + 6

Gii

y l mt h cn kh nng bin i tng i tt.T phng trnh u ta thy ngay

2x(1 y3) = 3(1 y2)

TH1 : y = 1 thay vo (2) ta c

x3 7x+ 6 = 0 x = 1, x = 3, x = 2

TH2 : Kt hp vi phng trnh (2) ta gy dng mt h mi{2x+ 2xy + 2xy2 = 3 + 3y()2x2 x3y = 2x2y2 7xy + 6

Nhng m phng trnh (2) li tng ng : (xy 2)(2xy + x2 3) = 0Sao m phn tch hay th ? Casio thn chng chng. C th, nhng ta hy vit li phngtrnh (2) mt cht

(2) 2x2y2 + xy(x2 7) 2x2 + 6 = 0xy =

(x2 7)2 8 (2x2 + 6) = (x2 + 1)2

Thy ri ch ? Coi xy l n chnh, tnh ra c kt qu m mn v t c hng phntch nhn t nh trnTH1 : xy = 2 thay li (*) ta c

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

y(1 + y) = 4Phng trnh ny v nghim nn trng hp 1 v nghimTH2 : 2xy = 3 x2 thay li (*) ta c

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

2x(

2

x 1)

= 3 x2 x = 1, y = 1

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


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Cu 133

{x (x+ y) +

x+ y =

2y(

2y3 + 1)

x2y 5x2 + 7 (x+ y) 4 = 6 3xy x+ 1

Gii

iu kin : y 0, x+ y 0Xut pht t phng trnh u, s dng phng php lin hp

PT (1) x2 + xy 2y2 =

2y x+ y

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

x+ y

R rng x+ 2y = x+ y + y > 0,1

2y +x+ y

< 0 nn t ta suy ra x = y

Thay vo phng trnh (2) ta c

x3 5x2 + 14x 4 = 6 3x2 x+ 1

y l mt loi phng trnh v t kh quen thuc. Cch gii tt nht vn l thm bt v xthm. Tuy nhin nu ca ta l thm bt x2x+ 1 vo 2 v xt hm t3 + 6t c v khngthnh cng v v tri khng phn tch c v dng . Ta hy kho lo bin i mt cht nhsauPhng trnh cho tng ng

x3 + 3x2 + 6x+ 4 = 8x2 8x+ 8 + 3 3

8x2 8x+ 8

(x+ 1)3 + 3 (x+ 1) = 8x2 8x+ 8 + 3 3

8x2 8x+ 8Nhn thy hm cn xt ri ch ? f(t) = t3 + 3t v hm n iu tng. T ta c

x+ 1 = 3

8x2 8x+ 8 x = 1, y = 1


S dng lin hp cng l phng php kh th v. Nu ta s dng n tt trong gii phngtrnh v t ri th khi i mt vi h phng trnh, ch cn mt cht nhn xt hnh thc cah v cc k nng tung ra, c th ta s thnh cng. Hy cnh gic vi nhng bi h m mtphng trnh cha nhiu cn thc, c th lin hp s l n nh tt nht chm p n

Tip theo ta n vi mt cu h s dng lin hp kh kh. Mong bn c th li v ti khngth din t ni v sao ti li lm vy. Mt kinh nghim khi tnh gii hn ca hm s gipti gii quyt c n.


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NguynMinhTun

2.5 Cu 121 n cu 150 73

Cu 134

{ x2 x y = y

3x y

2 (x2 + y2) 32x 1 = 11

Gii

iu kin : x 6= y, x 12, x2 x y 0

Phng trnh u tng ngx2 x y. 3x y = y

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

x2 x y (x y 1)

3

(x y)2 + 3x y + 1

+x2 x y y2x2 x y + y = 0

(x y 1) x2 x y

3

(x y)2 + 3x y + 1

+x+ y

x2 x y + y

= 0Thnh qu cng c cht t ri. Gi y ta ch mong em trong ngoc lun dng hoc mT phng trnh (1) ta thy ngay y v 3

x y phi cng du.

Gi s y < 0 th suy ra x y < 0 x < y < 0. R rng v l v iu kin l x 12.

Nh vy suy ra y > 0 x > y > 0 v hin nhin ngi p trong ngoc s lun dngTh pho nh nhm c ri. Gi hng th thnh qu ! Vi y = x 1 thay vo (2) ta c

2(x2 + (x 1)2) 32x 1 = 11 x = 5

2 y = 3

2(TM)


(5

2;3

2

)

Cu 135

{x4 + 2xy + 6y (7 + 2y)x2 = 92yx2 x3 = 10

Gii


x4 7x2 + 9 2y(x2 x 3) = 0 (x2 x 3)(x2 + x 3) 2y (x2 x 3) = 0

TH1 : x2 x 3 = 0

x = 1

13

2 y = 79 +

13

36

tuyen tap 410 cau he phuong trinh

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