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1/25

TRAN STUNG---- & ----

TI LIU N THI I HC CAO NG

Na m 2011

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2/25

Trn STng H phng trnh nhiu n

Trang1

1. H phng trnh bc nht hai n

a x b y c a b a b

a x b y c 2 2 2 21 1 11 1 2 2

2 2 2

( 0, 0) + =

+ + + =

Gii v bin lun:

Tnh cc nh thc:a b

Da b

1 1

2 2

= , xc b

Dc b

1 1

2 2

= , ya c

Da c

1 1

2 2

= .

Ch :gii h phng trnh bc nht hai n ta c thdng cc cch gii bit nh:phng php th, phng php cng i s.

2. H phng trnh bc nht nhiu nNguyn tc chung gii cc h phng trnh nhiu n l khbt na v ccphng trnh hay h phng trnh c sn t hn. kh bt n, ta cng c th dng ccphng php cng i s, phng php th nhi vi h phng trnh bc nht hai n.

Bi 1. Gii cc h phng trnh sau:a)

x y

x y

5 4 3

7 9 8

=

=b)

x y

x y

2 11

5 4 8

+ =

=c)

x y

x y

3 1

6 2 5

=

=

d)( )

( )x y

x y

2 1 2 1

2 2 1 2 2

+ + =

=e)

x y

x y

3 216

4 35 3

112 5

+ =

=

f)x y

y

3 1

5x 2 3

=

+ =

Bi 2. Gii cc h phng trnh sau:

a)x y

x y

1 818

5 451

=

+ =

b)x y

x y

10 11

1 225 3

2

1 2

+ = +

+ =

+

c)x y x y

x y x y

27 327

2 345 48

1

2 3

+ = +

=

+

d)x y

x y

2 6 3 1 5

5 6 4 1 1

+ + = + =

e)x y x y

x y x y

2 9

3 2 17

+ = + + =

f)x y x y

x y x y

4 3 8

3 5 6

+ + = + =

Bi 3. Gii v bin lun cc h phng trnh sau:a)

mx m y m x my

( 1) 1

2 2

+ = + + =

b)mx m y

m x m y ( 2) 5

( 2) ( 1) 2

+ = + + + =

c)m x y m m x y m

( 1) 2 3 1

( 2) 1

+ = + =

d)m x m y m x m y m

( 4) ( 2) 4

(2 1) ( 4)

+ + = + =

e)m x y m

m x y m m 2 2( 1) 2 1

2

+ =

= +f)

mx y m x my m

2 1

2 2 5

+ = + + = +

Bi 4. Trong cc h phng trnh sau hy:i) Gii v bin lun. ii) Tm m Z h c nghim duy nht l nghim nguyn.

a)m x y m

m x y m m 2 2( 1) 2 1

2

+ =

= +b)

mx yx m y m

1

4( 1) 4

= + + =

c)mx yx my m

3 3

2 1 0

+ = + + =

I. H PHNG TRNH BC NHT NHIU N

Xt D Kt qu

D 0 H c nghim duy nht yxDD

x yD D

;

= =

Dx 0 hoc Dy 0 H v nghimD = 0Dx = Dy = 0 H c v snghim


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H phng trnh nhiu n Trn STng

Trang2

Bi 5. Trong cc h phng trnh sau hy:i) Gii v bin lun.ii) Khi h c nghim (x; y), tm h thc giax,yc lp i vi m.

a)mx y m

x my m2 1

2 2 5

+ = + + = +

b)mx m y

m x my6 (2 ) 3

( 1) 2

+ = =

c)mx m y m

x my( 1) 1

2 2

+ = + + =


ax y bx y3 2 5

+ = + =

b)y ax b

x y2 3 4 = =

c)ax y a bx y a2

+ = + + =

d)a b x a b y a

a b x a b y b( ) ( )

(2 ) (2 )

+ + = + + =

e) ax by a bbx ay ab

2 2

2

+ = ++ =

f)ax by a b

bx b y b

2

2 4

=

=Bi 7. Gii cc h phng trnh sau:

a)

x y zx y z

x y z

3 1

2 2 5

2 3 0

+ = + =

=b)

x y zx y zx y z

3 2 8

2 6

3 6

+ + =+ + =

+ + =c)

x y zx y z

x y z

3 2 7

2 4 3 8

3 5

+ = + + =

+ =



4/25


Trang3

1. H gm 1 phng trnh bc nht v 1 phng trnh bc hai T phng trnh bc nht rt mt n theo n kia.

Th vo phng trnh bc hai a v phng trnh bc hai mt n. S nghim ca h tu theo s nghim ca phng trnh bc hai ny.

2. Hi xng loi 1

H c dng: (I) f x yg x y

( , ) 0

( , ) 0

= =

(vif(x, y) = f(y, x) v g(x, y) = g(y, x)).

(C ngha l khi ta hon v giax vy thf(x, y) v g(x, y) khng thay i).t S = x + y, P = xy.a h phng trnh (I) v h (II) vi cc n l Sv P. Gii h (II) ta tm c S v P.

Tm nghim (x, y) bng cch gii phng trnh: X SX P 2 0 + = .

3. Hi xng loi 2

H c dng: (I) f x yf y x( , ) 0 (1)

( , ) 0 (2)

= =

(C ngha l khi hon v giax vy th (1) bin thnh (2) v ngc li). Tr (1) v (2) v theo v ta c:

(I) f x y f y x f x y( , ) ( , ) 0 (3)

( , ) 0 (1)

= =

Bin i (3) v phng trnh tch:(3) x y g x y ( ). ( , ) 0 = x y

g x y( , ) 0

= =

.

Nh vy: (I)

f x y

x y

f x yg x y

( , ) 0

( , ) 0

( , ) 0

= = = =

.

Gii cc h trn ta tm c nghim ca h (I).Ch : Vi cc h phng trnh i xng, nu h c nghim x y0 0( ; ) th y x0 0( ; )

cng l nghim ca h. Do nu h c nghim duy nht th x y0 0= .

4. Hng cp bc hai

H c dng: (I)a x b xy c y d

a x b xy c y d

2 21 1 1 1

2 22 2 2 2

+ + =

+ + =.

Gii h khix = 0 (hocy = 0). Khix 0, t y kx= . Th vo h (I) ta c h theo kvx. Khx ta tm c phngtrnh bc hai theo k. Gii phng trnh ny ta tm c k, t tm c (x; y).

II. H PHNG TRNH BC HAI HAI N


5/25


Trang4

Bi 1. Gii cc h phng trnh sau:a) x y

x y

2 24 8

2 4

+ =+ =

b) x xyx y

2 24

2 3 1

= =

c) x yx y

2( ) 49

3 4 84

=+ =

d) x xy y x y x y

2 23 2 3 6 0

2 3

+ + + = =

e)x y

xy x y 3 4 1 0

3( ) 9

+ = = +

f)x y

xy x y 2 3 2

6 0

+ = + + + =

g) y x x

x y

2 4

2 5 0

+ =+ =

h)x y

x y y2 22 3 5

3 2 4

+ =

+ =i)

x y

x xy y 2 22 5

7

=

+ + =


x y

x y m2 26 + =

+ =

b)x y m

x y x2 2 2 2

+ =

+ =c)

x y

x y m2 23 2 1 =

+ =

Bi 3. Trong cc h phng trnh sau:i) Tm s nguyn m h c nghim duy nht l nghim nguyn.ii) Khi h c nghim (x, y) , tm h thc giax, yc lp vi m.

a)

mx y m

x my a

2 1

2 2 1

+ = + + = b)

mx y m

x my m

3

2 1

+ = + = +

c)x y m

x y m2 4

2 3 3

= + = +

d)x y

y x m2 5

2 10 5

+ = = +


x xy y

x y xy x y 2 211

2( ) 3

+ + =

+ + = b)

x y

x xy y 2 24

13

+ =

+ + =c)

xy x y

x y x y 2 25

8

+ + =

+ + + =

d)

x y

y x

x y

13

6

6

+ =

+ =

e) x x y y x y xy

3 3 3 3 17

5

+ + =+ + =

f)x x y y

x xy y

4 2 2 4

2 2

481

37

+ + =

+ + =


x xy y

x y y x 2 21

6

+ + =

+ = b)

x y

x x y y

2 2

4 2 2 4

5

13

+ =

+ =c)

x y y x

x y

2 2

3 3

30

35

+ =

+ =

d)x y

x y x y

3 3

5 5 2 2

1 + =

+ = +e)

x y xy

x y x y

2 2

4 4 2 2

7

21

+ + =

+ + =f)

x y xy

x y x y 2 211

3( ) 28

+ + =

+ + + =

Bi 6. Gii cc h phng trnh sau:

a)

x yxy

x yx y

2 2

2 2

1( ) 1 5

1( ) 1 49

+ + =

+ + =

b) ( )

y x x y

x yx y

2 2

2 2

2 2

( 1) 2 ( 1)

11 24

+ = +

+ + =

c)

x yx y

x yx y

2 2

2 2

1 14

1 14

+ + + =

+ + + =

d)

x y

x y

x yxy

2 2

2

31 11

( )(1 ) 6

+ = + +

+ + =

e)x y y x y x xy

y xxy xy x y

2 22 2 6

1 4

+ + + =

+ + + =

f)

xyxy

x yxy

14

1( ) 1 5

+ =

+ + =


6/25


Trang5


x y xy m

x y m2 2 3 2

+ + =

+ = b)

x y m

x y xy m m 2 2 21

2 3

+ = +

+ = c)

x y mxy x y m ( 1)( 1) 5

( ) 4

+ + = + + =


x x y

y y x

2

2

3 2

3 2

= +

= +b)

x y x y

y x y x

2 2

2 2

2 2

2 2

= +

= +c)

x x y

y y x

3

3

2

2

= +

= +

d)

yx y

xx

y xy

3 4

3 4

=

=

e)

yy

x

xx

y

2

2

2

2

23

23

+=

+ =

f)

x yy

y xx

2

2

12

12

= +

= +

g)x x y

y y x

3

3

3 8

3 8

= +

= +h)

xy x y

xy y x

2

2

8( 1)

8( 1)

+ =

+ = i)

x yx

y x

y

2

2

32

32

+ =

+ =

k)

2 2

2 2

91 2 (1)

91 2 (2)

+ = +

+ = +

x y y

y x x


x x my

y y mx

2

2

3

3

= +

= +b)

x y m m

y x m m

2 2

2 2

(3 4 ) (3 4 )

(3 4 ) (3 4 )

=

= c)

xy x m y

xy y m x

2

2

( 1)

( 1)

+ =

+ =

Bi 10. Tm m h phng trnh sau c nghim duy nht:

a) x y m y xy m x

2 2

2 2 + =

+ =b) xy x m y

xy y m x

2

2( 1)( 1)

+ = + =

c)

m

x y y

my x

x

22

22

2

2

= + = +


x xy y

x xy y

2 2

2 2

3 1

3 3 13

+ =

+ =b)

x xy y

x xy y

2 2

2 2

2 4 1

3 2 2 7

+ =

+ + =c)

y xy

x xy y

2

2 2

3 4

4 1

=

+ =

d)x xy y

x xy y

2 2

2 2

3 5 4 38

5 9 3 15

+ =

=e)

x xy y

x xy y

2 2

2 2

2 3 9

4 5 5

+ =

+ =f)

x xy y

x xy y

2 2

2 2

3 8 4 0

5 7 6 0

+ =

=


x mxy y m

x m xy my m

2 2

2 2( 1)

+ + =

+ + =b)

xy y

x xy m

2

2

12

26

=

= +c)

x xy y m

y xy

2 2

2

4

3 4

+ =

=


7/25


Trang6

Cc h phng trnh i s tng qut thng rt kh gii v khng th nu ra phngphp chung gii chng. y xin nu ra mt s phng php c th la chn thchhp.

1. Phng php th: T phng trnh n gin nht ca h hoc t phng trnh tchtm cch rt mt n theo n kia, ri th vo phng trnh cn li. Gii phng trnh ny.S nghim ca h tu thuc s nghim ca phng trnh ny.2.t n ph: Bin i cc phng trnh c tht n ph, ri chuyn v h cbn.3.Phng php nh gi: Tiu kin ca n, xt trng hp xy ra du "=" btng thc.4. Phng php iu kin cn v :5. Phng php hm s: Chn hm s thch hp, ri s dng tnh n iu ca hm s.

Cho hm sy = f(x) ng bin (hoc nghch bin) trn khong (; ). Khi , vi mia, b (; ) ta c: f(a) = f(b) a = b.

Ch : Cc h phng trnh hon v vng quanhx f yy f zz f x

( )

( )

( )

= =

=, thng sdng tnh n

iu ca hm schng minh x = y = z. Xt tnh n iu hm sf(t). Chng t x < y, x > y, khng xy ra. T suy ra x = y = z. Thvo h cho gii tm x, y, z.

Vn 1: Phng php th

Bi 1. Gii h phng trnh sau:x y y x y

x y x y

2

2

1 ( ) 4 (1)

( 1)( 2) (2)

+ + + =

+ + =

Dthy y 0. HPT[ ]y y y x y x y 4 ( ) ( 2) + + = [ ]y x2

(3 ) 0 = y x3= Nghim: (1; 2), (2; 5).

Bi 2. Gii h phng trnh sau:2 2

2 2

3 (1)

1 1 4 (2)

+ = + + + =

x y xy

x y

(2) x y x y xy xy xy 2 2 2 2 22 ( 1).( 1) 14 2 ( ) 4 11+ + + + = + + + = (3)

t xy = p.p

pp p p

pp p

22

311

(3) 2 4 11 353 26 105 0

3

= + + = =+ =

(1) ( )x y xy 2

3 3+ = + p = xy =35

3 (loi) p = xy = 3 x y 2 3+ =

1/ Vixy

x yx y

33

2 3

=

= = + =2/ Vi

xyx y

x y

33

2 3

=

= = + =

Vy h c hai nghim l: ( ) ( )3; 3 , 3; 3

III. H PHNG TRNH DNG KHC


8/25


Trang7

Bi 3. Gii h phng trnh sau:x x y

x yx

2

2

( 1) 3 0

5( ) 1 0

+ + = + + =

(D 2009)

V x 0 nn HPTx y

x

x yx

2

2

31

5( ) 1 0

+ =

+ + =

x y

x

xx2

31

4 62 0

+ =

+ =

xxx y x y

1 11

1 21

22

= = + = + =

. Nghim:3

(1;1), 2;2

.

Bi 4. Gii h phng trnh sau:x x y xy y

x y x y

3 2 2 36 9 4 0 (1)

2 (2)

+ =

+ + =

Ta c: (1) x y x y 2( ) ( 4 ) 0 = x yx y4

= =

Vi x = y: (2) x = y = 2

Vi x = 4y: (2) x y32 8 15; 8 2 15= = Bi 5. Gii h phng trnh sau:

+ + =

+ + + =

x x y

x x y xy x

2

3 2 2

5 9

3 2 6 18

H y x x

x x x x+

2

4 3 2

9 5

4 5 18 18 0

=

+ =

y x xxx

x

2

9 51

3

1 7

= = = =

x yx y

x y

x y

1; 33; 15

1 7; 6 3 7

1 7; 6 3 7

= = = =

= = + = + =

Bi 6. Gii h phng trnh sau:x y xy

x y

2 0

1 4 1 2

=

+ =

H PT( ) ( )x y x y

x y

2 0

1 4 1 2

+ =

+ =

x y

x y

2 0

1 4 1 2

=

+ =

x y

y

4

4 1 1

= =

x

y

2

1

2

= =

Bi 7. Gii h phng trnh sau:xy

x yx y

x y x y

2 2

2

21 (1)

(2)

+ + =

+ + =

iu kin: x y 0+ > .

(1) x y xy x y

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

(v x y 0+ > nn x y x y 2 2 0+ + + > )


9/25


Trang8

Thay x y1= vo (2) ta c: x x21 (1 )= x x2 2 0+ = x yx y

1 ( 0)

2 ( 3)

= = = =

Vy h c 2 nghim: (1; 0), (2; 3).Bi 8. Gii h phng trnh sau:

( )x y xy

x y

3 3

2 2

3 4 (1)

9 (2)

=

=

T(2): x y xy 2 2 9 3= = .

Khi: xy 3= , ta c: x y3 3 4 = v ( )x y3 3. 27 =

Suy ra: ( )x y3 3; l cc nghim ca phng trnh: X X X2 4 27 0 2 31 = =

Vy nghim ca H PT l x y3 32 31, 2 31= + =

hoc x y3 32 31, 2 31= = + .

Khi: xy 3= , ta c: x y3 3 4 = v ( )x y3 3. 27 =

Suy ra: x y3 3; ( ) l nghim ca phng trnh: X X PTVN 2 4 27 0 ( )+ + = Bi 9. Gii h phng trnh sau:

( )2

3 2 (1)

2 8 (2)

=

=

x y xy

x y

iu kin : x y x y . 0 ;

Ta c: (1) x y xy x y x y 23( ) 4 (3 )( 3 ) 0 = =y

x y hay x 33

= =

Vi x y3= , thvo (2) ta c : y y y y 2 6 8 0 2 ; 4 + = = =

H c nghim x xy y

6 12;

2 4

= = = =

Viy

x3

= , thvo (2) ta c : y y23 2 24 0 + = V nghim.

Kt lun: h phng trnh c 2 nghim l:x xy y

6 12;

2 4

= = = =

Bi 10.Gii h phng trnh sau:x y y x

y x

3 3

2 2

4 16 (1)

1 5(1 ) (2)

+ = +

+ = +

T(2) suy ra y x2 2 5 4= (3). Thvo (1) c:

( )yx x y y x 2 23 3 5 . 16+ = + x x y x 3 2 5 16 0= xx xy2

0

5 16 0

= =

Vi x 0= y2 4= y 2= .

Vi x xy2 5 16 0= xyx

2 16

5

= (4). Thvo (3) c:

x

xx

22

216

5 45

= x x x x 4 2 4 2

32 256 125 100+ =

x x4 2124 132 256 0+ = x2 1= x yx y

1 ( 3)1 ( 3)

= = = =

.


10/25


Trang9

Vy h c 4 nghim: (x; y) = (0; 2) ; (0; 2); (1; 3); (1; 3)Bi 11.Gii h phng trnh sau:

xy y

xy x

2

2

4 8

2

=

= +

Nu xy 4 th HPT

xy y

xy x

2

2

4 8 (1)

2 (2)

=

= +

T(2) x 0, x2 2 vx

yx

22 +=

Thay vo (1) ta c:x

xx

22

2 22 4 8 +

+ =

x x2 2( 2)( 1) 0 = x 2=

H c nghim (x; y) l: ( ) ( )2; 8 , 2; 8

Nu xy < 4 th x2 2< .

HPT xy yxy x

2

24 8

2 =

= + xx

x

2

22 24 2 8 + =

x22(2 ) 0 = x2 2= (loi)

Kt lun: Nghim (x; y) ca h: ( ) ( )2; 8 , 2; 8 Bi 12.Gii h phng trnh sau:

x y x y x x

xy x x

2 2

2

( 1)( 1) 3 4 1 (1)

1 (2)

+ + + = +

+ + =

T(2) x 0 vx

yx

2 11

+ = . Thay vo (1) ta c:

x xx x x x

x x

2 22 21 1 3 4 1

+ = +

x x x2 ( 1)( 2) 0 + = xx

1

2

= =

(v x 0)

Nghim (x; y):5

(1; 1), 2;2

Bi 13.Gii h phng trnh sau:xy x y x y

x y y x x y

2 22 (1)

2 1 2 2 (2)

+ + =

=

iu kin x 1, y 0 x + y > 0.(1)

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

x y2 1= +

(3)

Thay (3) vo (2) ta c: y y y y y y (2 1) 2 2 2(2 1) 2+ = +

( )y y( 1) 2 2 0+ = y = 2 x = 5Nghim (x; y): (5; 2)

Bi 14.Gii h phng trnh sau:y x x

y x xy x y

2

2 2

(5 4)(4 ) (1)

5 4 16 8 16 0 (2)

= +

+ + =

T(1) y x x2 25 16 16= + + .

Thay vo (2) ta c: y xy y 22 4 8 0 = yy x02 4 = = +


11/25


Trang10

Vi y = 0 x x25 16 16 0 + + = x

x

4

54

=

=

Vi y x2 4= + x x x2 2(2 4) 5 16 16+ = + + x = 0 y = 4.

Kt lun: Nghim (x; y):

4

(0; 4), (4; 0), ; 05

.Bi 15.Gii h phng trnh sau:

xyx y

x y

x y x y x

2 2

3

816 (1)

3 3 2 1 (2)

+ + =

+ + + + =

Dthy x + y > 0.

T(1) x y xy x y

2 4( ) 16 2 1 0

+ = + x y x y x y 2 2( 4) 4( ) 0 + + + + =

x y 4 0+ = x y 4+ =

Thay vo (2) ta c: x x3 2 7 3 2+ = x x x3 22 9 14 5 0 + = x y1 7

2 2

= =

Nghim (x; y):1 7

;2 2

.

Bi 16.Gii h phng trnh sau: x x y y x y x y

4 2 2

2 2

4 6 9 0 (1)

2 22 0 (2)

+ + =

+ + =

T(2)x

yx

2

2

22

2

=

+. Thay vo (1) ta c:

xx x

x

224 2

2

224 3 0

2

+ =

+

xx x

x

2 22 2

2 2

16( 4)( 4) 0

( 2)

+ =

+

x x x x 2 6 4 2( 4)( 4 20 64) 0 + + =

x yx y

x y

x y

2 ( 3)

2 ( 3)

2 ( 5)

2 ( 5)

= = = =

= = = =

Bi 17.Gii h phng trnh sau:x y x y

x y x y

3 (1)

2 (2)

=

+ = + +(B - 2002)

iu kin:x y

x y0

(3)0

+

(1) ( ) x yx y x y x y

3 61 01

= = = +. Thay vo (2) ta c:

x y

x y

1

3 1,

2 2

= =

= =

.

Bi 18.Gii h phng trnh sau:x y

x y

y x3

1 1(1)

2 1 (2)

=

= +

(A - 2003)

iu kin xy 0. Ta c: (1) x yx yxyxy

1( ) 1 0

1

= + = =


12/25


Trang11

Trng hp 1:

x y

x y x y x y

x x x x x

x y

3 2

1

1 5

22 1 ( 1)( 1) 0

1 5

2

= = = = + = =

= + + =

= =

Trng hp 2:yxy yx

xy x

x x x VN x

33 4

1 11

22 11 2 0 ( )

= = = = + = + + + =

Kt lun: Nghim (x; y):1 5 1 5 1 5 1 5

(1;1), ; , ; ;2 2 2 2

+ +

.

Bi 19.Gii h phng trnh sau:x x y y

x y

3 3

2 2

8 2

3 3( 1)

= +

= +

(DB A 2006)

H PT x y x y

x y

3 3

2 2

3( ) 6(4 ) (1)

3 6 (2)

= +

=.

Th(2) vo (1) ta c: x y x y x y 3 3 2 23( ) ( 3 )(4 ) = + x x y xy 3 2 212 0+ =

xx yx y

0

3

4

= = =

.

Nghim (x; y):6 6 6 6

(3;1), ( 3; 1), 4. ; , 4. ;13 13 13 13

.

Bi 20.Gii h phng trnh sau:


13/25


Trang12

Vn 2: Phng php t n ph

Bi 1. Gii h phng trnh sau:x x y y

x y x y

4 2 2

2 2

4 6 9 0

2 22 0

+ + =

+ + =

HPT x yx y x

2 2 2

2 2( 2) ( 3) 4

( 2 4)( 3 3) 2 20 0 + =

+ + + =. t u x

v y

2 23

= =

.

HPT u uu vv vuv u v

2 2 2 040 24( ) 8

= =+ = = =+ + = . Nghim (2; 3), ( 2; 3),( 2; 5), ( 2; 5)

Bi 2. Gii h phng trnh sau:x y y

x y x y

3 3 3

2 2

8 27 18

4 6

+ =

+ =

HPTx

y

x xy y

33 3(2 ) 18

3 32 . 2 3

+ =

+ =

. t a = 2x; b =y3 . HPT a b

ab3

1 + = =

H cho c nghim:3 5 6 3 5 6

; , ;4 43 5 3 5

+ +

Cch 2: D thy y 0. HPT x y y

x y xy y

3 3 3

2 2 3

8 27 18

4 6

+ =

+ = x y x y xy 3 3 2 28 27 18(4 6 )+ = +

(*)

t t xy= . (*) t t t2(2 3)(4 42 9) 0+ + = t

t

32

21 9 5

4

=

=

Bi 3. Gii h phng trnh sau:x y y x y

x y x y

2

2

1 ( ) 4

( 1)( 2)

+ + + =

+ + =

Dthy y 0. HPT

xy x

y

xy x

y

2

2

12 2

1( 2) 1

++ + =

+ + =

.t

xu

yv y x

2 1

2

+ = = +

.

HPT u vuv

2

1

+ = =

uv

1

1

= =

x

y

y x

2 11

2 1

+=

+ =

x

y

1

2

=

=hoc

x

y

2

5

=

=

Bi 4. Gii h phng trnh sau: + + =

+ + + =

x x y

x x y xy x

2

3 2 2

5 9

3 2 6 18

HPT x x x y x x x y

2

2

2 3 9

( 2 )(3 ) 18

+ + + =

+ + =. t u x x

v x y

2 2

3

= += +

.


14/25


Trang13

HPT u vuv

9

18

+ = =

u vu v

6, 3

3, 6

= = = =

. Nghim:

x yx y

x y

x y

1; 3

3; 15

1 7; 6 3 7

1 7; 6 3 7

= = = =

= = + = + =

Bi 5. Gii h phng trnh sau:xy x y x y xy y 2 2 2

1 71 13

+ + =+ + =

(B - 2009)

Dthy y 0. HPT

xx

y y

xx

y y

2

17

113

+ + =

+ =

.tu x

yx

vy

1= +

=

.

HPTu v

u v27

13

+ =

= u

v5

12

= =

hoc uv

4

3

= =

. Nghim1

1; , (3;1)3

.


3 3

2 1

2 2

y x

x y y x

=

=

HPT ( )( )x y y x y x x x y xy y 3 3 2 2 3 2 2 32 2 2 2 2 5 0 = + + = Khi y 0= th h VN.

Khi y 0 , chia 2 vcho y3 0 ta c:x x x

y y y

3 2

2 2 5 0

+ + =

tx

t y= , ta c : t t t t 3 22 2 5 0 1+ + = =

y xx yx yy2 1 11 = = = = = =

Bi 7. Gii h phng trnh sau:y

xx yx

x yy

2 2

2 2

32 1

1

4 22

+ = +

+ + =

iu kin: x y x y 2 20, 0, 1 0 +

tx

u x y v y2 2 1;= + = . H PT trthnh: u v u v

u v u v

3 2 3 2

1 1 (1)1 4 22 21 4 (2)

+ = + = + + = =

Thay (2) vo (1) ta c:v

v vvv v

23

3 21 2 13 21 0 7

21 42

=+ = + =

=

Nu v = 3 th u = 9, ta c H PT:

x yx xx y

xy yx y

y

2 22 21 9 3 310

1 13 3

+ = = = + = = = = =

Nu v 72

= th u = 7, ta c H PT:


15/25


Trang14

y yx y x yx

x yy x x

2 2 2 2 2 24 41 7 853 537 7

2 22 14 142

53 53

= = + = + = = = = =

So snh iu kin ta c 4 nghim ca H PT.


2 2

1 4

( ) 2 7 2

x y xy y

y x y x y

+ + + =

+ = + +

Th PT y 0 . Khi ta c:

xx y

x y xy y y

y x y x y x x y

y

2

2 2

2 2 22

14

1 4.

( ) 2 7 2 1( ) 2 7

++ + = + + + =

+ = + + + + =

tx

u v x y

y

2 1,

+= = + ta c h:

u v u v v u

v uv u v v 2 2

4 4 3, 1

5, 92 7 2 15 0

+ = = = = = = = + =

Vi v u3, 1= = ta c h:x yx y x y x x

y x x yx y y x

2 2 2 1, 21 1 2 0

3 2, 53 3

= = + = + = + = = = =+ = = .

Vi v u5, 9= = ta c h: x y x y x x x y y x y x

2 2 21 9 1 9 9 46 0

5 5 5

+ = + = + + = + = = =

, h VN

Kt lun: H cho c hai nghim: (1; 2), ( 2; 5) .Bi 9. Gii h phng trnh sau:

=++++

=++++

011)1(

030)2()1(

22

3223

yyyxyx

xyyyxyyx

H PT xy x y x y x y xy x y xy x y

2 2 2( ) ( ) 30

( ) 11

+ + + =+ + + + =

xy x y x y xy xy x y xy x y

( )( ) 30

( ) 11

+ + + = + + + + =

tx y uxy v

+ = =

. HPTuv u v uv u v

( ) 30

11

+ = + + =

uv uv uv u v

(11 ) 30 (1)

11 (2)

= + + =

. T(1)uvuv

5

6

= =

Vi uv = 5 u v 6+ = . Gii ra ta c cc nghim (x; y) l: 5 21 5 21;2 2

+

v

5 21 5 21;

2 2

+

Vi uv = 6 u v 5+ = . Gii ra ta c cc nghim (x; y) l: (1;2) v (2;1)

Kt lun: H PT c 4 nghim: (1;2), (2;1),5 21 5 21

;2 2

+

,5 21 5 21

;2 2

+


x y x y

2 2

2 2

3 4 1

3 2 9 8 3

+ + =

=

HPT x x y y

x x y y

2 2

2 2

3 4 1

3( 3 ) 2( 4 ) 3

+ + =

+ =

.t u x x

v y y

2

2

3

4

=

= +

.

HPT u vu v

1

3 2 3

+ = =

. Nghim (x; y):3 13 3 13

; 0 , ; 42 2

.


16/25


Trang15


x y x y 2 2 2 2

2

1 3

+ = +

+ + =

iu kin: x y x y 0, 0+ > .t:u x y

v x y

= +

=

ta c h:

u v u v u v uv

u v u v uv uv

2 2 2 2

2 ( ) 2 4

2 23 3

2 2

= > + = +

+ + + + = =

u v uv

u v uv uv

2

2 4 (1)

( ) 2 23 (2)

2

+ = +

+ + =

.

Th(1) vo (2) ta c:uv uv uv uv uv uv uv

2

8 9 3 8 9 (3 ) 0+ + = + + = + =.

Kt hp (1) ta c:uv

u vu v

04, 0

4

= = =

+ =(vi u > v).

T ta c: x = 2; y = 2.(thok)Kt lun: Vy nghim ca h l: (x; y) = (2; 2).

Bi tng t:x y x y

x y x y 2 2 2 2

2

4

+ =

+ + =

Bi 12.Gii h phng trnh sau:xy x y

x y x y 2 2

3 2 16

2 4 33

=

+ =

HPTx y x y

x y2 2( 1)( 2) ( 1) ( 2) 21

( 1) ( 2) 38

=

+ =.t u x

v y1

2

= =

.

HPTuv u v

u v2 2( ) 21

38

+ =

+ =. Nghim (x; y): ( ) ( )3 3; 2 3 , 3 3; 2 3 + + .

Bi 13.Gii h phng trnh sau:x y xy

x y

xx y

2 2

2

34( ) 7

( )

12 3

+ + + = +

+ = +

HPTx y x y

x y

x y x y x y

2213 ( ) 13

13

+ + + = + + + + = +

.tu x y

x yv x y

1= + +

+ =

(vi u 2 )

HPT u vu v

2 23 13

3

+ =+ =

u

v uv

2( 2)

1

= =

x y xx y

y

x y

12 1

0

1

+ + = = + = =

.



17/25


Trang16

=+

=+

358

152

33

22

yx

xyyx

H PTxy x y

x y3 32 (2 ) 30

(2 ) 35

+ =

+ =.t

u xv y

2 = =

. H PTuv u v

u v3 3( ) 30

35

+ =

+ = u v

u v2; 3

3; 2

= = = =

Nghim (x; y): 3(1;3), ; 22 .


x y

2 1 1

3 2 4

+ + + =

+ =

H PT x y x y x y x y

2 1 1

(2 1) ( ) 5

+ + + =

+ + + + =.t u x y v x y 2 1 0, 0= + + = + .

H PTu v u v

u v loai u v2 21 2, 1

1, 2 ( )5

= = = = = + =

xy

2

1

= =

.


=++

=++

64

9)2)(2(

2 yxx

yxxx

H PT x x x y

x x x y

2

2

( 2 )(2 ) 9

( 2 ) (2 ) 6

+ + =

+ + + =.t u x x

v x y

2 2

2

= += +



18/25


Trang17

Vn 3: Phng php nh gi

Bi 1. Gii h phng trnh sau:x z z a

y x x b

z y y c

3 2

3 2

3 2

9 27( 1) ( )

9 27( 1) ( )

9 27( 1) ( )

= = =

Cng (a), (b), (c) ta c: x y z d 3 3 3( 3) ( 3) ( 3) 0 ( ) + + = + Nu x > 3 th t(b) suy ra: 3 9 ( 3) 27 27 3y x x y= + > >

t(c) suy ra: 3 9 ( 3) 27 27 3z y y z= + > > (d) khng tho mn+ Tng t, nu x < 3 th t(a) 0 < z < 3 0 < y 0, y > 0, z > 0.

Khng mt tnh tng qut, gi sx y y z y z 1 1+ + .

Ta li c: z x y x 1 1= + + = x y z x x = y = z.

x x 1 0 = ( )

x

25 1

4

+= . Nghim x = y = z =

( )2

5 1

4

+.

Bi 3. Gii h phng trnh sau:x

yx

yz

y

zx

z

2

2

2

2

2

2

2

1

2

1

2

1

= + =

+

= +

Nu x = 0 th y = 0, z = 0 H c nghim (x; y; z) = (0; 0; 0) Nu x 0 th y > 0, z > 0 x > 0.

Ta c:x x

y x

xx

2 2

2

2 2

21

= =

+

. Tng tta suy ra c: y x z y x = y = z

x

xx

2

2

2

1=

+ x = 1. Nghim (0; 0; 0), (1; 1; 1).

Bi 4. Gii h phng trnh sau:xy

x x yx x

xyy y x

y y

2

3 2

2

23

2

2 92

2 9

+ = +

+ + = + +

Cng hai phng trnh, vtheo v, ta c:xy xy

x yx x y y

2 2

3 2 23

2 2

2 9 2 9+ = + + + (*)

Ta c: x x x3 2 232 9 ( 1) 8 2 + = + , y y y2 23 32 9 ( 1) 8 2 + = +


19/25


Trang18

VT (*) xy xy

xy xy x y 2 22 2

2 22 2

+ = +

Du "=" xy ra x yx y

1

0

= = = =

. Nghim: (0; 0), (1; 1).

Bi 5. Gii h phng trnh sau:y x xx y y

3

33 4

2 6 2 = + +

=

y x x

x y y

3

3

3 4

2 6 2

= + +

= y x x

x y y

2

2

2 ( 1) ( 2) (1)

2 2( 1) ( 2) (2)

= +

= +

Ddng thy (x; y) = (2; 2) l mt nghim ca h.Nu x > 2 th t(1) y < 2. Nhng t(2) x 2 v y 2 cng du Mu thun.Nu x < 2 th cng suy ra iu mu thun tng t.Vy h c nghim duy nht x = y = 2.

Bi 6. Gii h phng trnh sau:xy x

xy y

2

2

10 20 (1)

5 (2)

= = +

HD : Rut ra yyy

yx +=

+=

55 2

C si202 x theo (1) 202 x suy ra x,y

T(2) yyy

yx +=

+=

55 2.

p dng BT C-si ta c: 525

+= yyx x2

20 . M theo (1) th x2

20 .

Do x2 20= x yx y

2 5 ( 5)

2 5 ( 5)

= =

= =

Bi 7. Gii h phng trnh sau:


20/25


Trang19

Vn 4: Phng php sdng phng trnh h qu

Bi 1. Gii h phng trnh sau:x yy z

z x

1 (1)

2 (2)

3 (3)

+ =+ =

+ =

Cng 3 phng trnh, vtheo v, ta c: x y z 3 (4)+ + = T(4) v (1) z = 2; t(4) v (2) x = 1; t(4) v (3) y = 0.Thli Nghim (x; y; z): (1; 0; 2).

Bi 2. Gii h phng trnh sau:xy x y yz y z zx z x

2 1

2 7

2 2

= + += + +

= + +

xy x y yz y z

zx z x

2 1

2 7

2 2

= + += + +

= + +

x yy z

z x

(2 1)(2 1) 3

(2 1)(2 1) 15

(2 1)(2 1) 5

= =

=

Nhn cc phng trnh trn, vtheo v, ta c:

x y z2 2 2(2 1) (2 1) (2 1) 225 = x y z a x y z b

(2 1)(2 1)(2 1) 15 ( )

(2 1)(2 1)(2 1) 15 ( )

= =

Trng hp (a) x

yz

2 1 1

2 1 3

2 1 5

= =

=

x

yz

1

2

3

==

=

Trng hp (b) xy

z

2 1 1

2 1 3

2 1 5

= =

=

xy

z

0

1

2

==

=

Thli Nghim (x; y; z): (1;2;3), ( 0; 1; 2) .Bi 3. Gii h phng trnh sau:


21/25


Trang20

Vn 5: Phng php hm s

Bi 1. Gii cc h phng trnh sau:y

x

x x x

y y y

2 1

2 1

2 2 3 1

2 2 3 1

+ + = +

+ + = +

tu xv y

1

1

= =

. HPTv

u

u u

v v

2

2

1 3

1 3

+ + =

+ + =

u vu u v v 2 23 1 3 1+ + + = + + + f u f v ( ) ( )= vi tf t t t 2( ) 3 1= + + + .

Ta c: tt t

f t

t

2

2

1( ) 3 ln 3 0

1

+ + = + >+

f(t) ng bin.

u v= ( )uu u u u u 2 231 3 log 1 0+ + = + + = (2)

Xt hm s: ( )g u u u u 23( ) log 1= + + g u( ) 0 > g(u) ng bin.M g(0) = 0 nn u = 0 l nghim duy nht ca (2).Nghim: (1; 1).

Bi 2. Gii h phng trnh sau:x x y y

x y

3 3

8 4

5 5 (1)

1 (2)

=

+ =

T(2) x y8 41, 1 x y1, 1 .

Xt hm s f t t t t 3( ) 5 , [ 1;1]= f t t t 2( ) 3 5 0, [ 1;1] = < f(t) nghch bin

trn [1; 1].Do : T(1) f(x) = f(y) x = y.

Thay vo (2) ta c: x x8 4 1 0+ = x y41 5

2

+= =

Bi tng t: x x y y

x y

3 3

6 6

3 3

1

=

+ =

Bi 3. Gii h phng trnh sau:


22/25


Trang21

Vn 6: Gii phng trnh bng cch a v h phng trnh

Bi 1. Gii phng trnh sau: 3 18 1 2 2 1++ = x xt x xu v3 12 0; 2 1+= > = .

Ta c h

u vu v u v

u uv u u v u uv v

3 3

33 2 2

01 2 1 2

2 1 01 2 ( )( 2) 0

= > + = + =

+ =+ = + + + =

x

x 2

0

1 5log

2

= + =

Bi 2. Gii phng trnh sau: x x3 31 2 2 1+ = t y x3 2 1= . Ta c h x y

y x

3

3

1 2

1 2

+ =

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

x x3 2 1 0 + = x

x

1

1 5

2

= =

.

Bi 3. Gii phng trnh sau: x x32 3 2 3 6 5 8 0 + = t u x v x v 3 3 2, 6 5 , 0 (* )= = .

Ta c h:u v

u v3 22 3 8

5 3 8

+ =

+ =

uv

u u u3 2

8 2

3

15 4 32 40 0

=

+ + =

u

v2

4

= =

x = 2.

Bi 4. Gii phng trnh sau:


23/25


Trang22

Vn 7: H phng trnh cha tham s

Bi 1. Tm m h phng trnh:( )

2 2

2 2

2

4

+ =

+ =

x y x y

m x y x yc ba nghim phn bit.

H PT

m x m x m

xy

x

4 2

2

2

( 1) 2( 3) 2 4 0 (1)

2

1

+ + =

+=+

.

+ Khi m = 1: H PTx

VNxy

x

2

2

2

2 1 0

( )2

1

+ = +

=+

+ Khi m 1.t t = x2 , t 0 . Xt f t m t m t m 2( ) ( 1) 2( 3) 2 4 0 (2)= + + =

H PT c 3 nghim phn bit (1) c ba nghim x phn bit

(2) c mt nghim t = 0 v 1 nghim t > 0 ( )f

mmS

m

(0) 0... 22 3

01

= == >

.

Bi 2. Tm m h phng trnh sau c nghim: 11 3

+ =

+ =

x y

x x y y m(D 2004)

t u x v y u v , ( 0, 0)= = . H PTu v u v

uv mu v m3 31 1

1 3

+ = + = =+ = .

S: m1

0

4

.

Bi 3. Tm m h phng trnh sau c nghim duy nht: y x my xy

2 (1)

1 (2)

= + =

T (1) = x y m2 , nn (2) = y my y 22 1 = +

y

m yy

1

12

(vy 0)

Xt ( ) ( )= + = + >f y y f y y y21 1

2 ' 1 0

Da vo BTT ta kt lun c h c nghim duy nht >m 2 .Bi 4. Tm m h phng trnh sau:

a)


24/25


Trang23


=+

=+

22

22

xy

yxb)

x y x y

x y

. 3

1 1 4

+ = + + + =

c)

x y

y x xy

x xy y xy

71

78

+ = +

+ =

d)

x y

y x

x y xy2 2

5

2

21

+ =

+ + =

e)x y

y x

x y xy

2 23

3

+ = + =

f)( )x y x y

x y x y

3 3

2 2

7

2

=

+ = + +

g) x yxy

1 1 43

9

+ = =

h) x y

xy

3 3 4

27

+ =

=


xy x y

x y x y xy2 23

6

+ =

+ + + =b) x xy y

x y xy

2 2 3

1

+ + =+ + =

c) x y

x y2 2

1 1 1

2

5

+ =

+ =

d)( ) ( )

x y x y

x x y y y

2 2 4

1 1 2

+ + + =

+ + + + =


x y x

y x y

2 2

2 2

2 3 2

2 3 2

=

= b)

x y x

y x y

3

3

2 2

2 2

= + +

= + +

c)x y x

y x y

2 2

2 2

2 3 4

2 3 4

= +

= +d)

x y x y

y x y x

2 2

2 2

2 2

2 2

= +

= +

e)

yx

yx

yx

2

2

2

12

1

=

=

f)

yx

y

xyx

2

2

2

2

1

1

11

=

+

= +Bi 4. Gii cc h phng trnh sau:

a)x xy y

x xy y

2 2

2 2

6 2 56

5 49

=

=b)

x xy y

x xy y

2 2

2 2

2 3 15

2 8

+ + =

+ + =

c)x xy y

x xy y

2 2

2 2

2 3 9

2 2 2

+ + =

+ + =d)

x xy y

x xy y

2 2

2 2

2 4 1

3 2 2 7

+ =

+ + =

e)x y

x y

5 5

3 3

1

1

+ =

+ =

f)x y

x y

2 2

3 3

1

1

+ =

+ =

IV. BI TP N



25/25


g)x y

x x y y

2 2

4 2 2 4

5

13

+ =

+ =h)

x y

x y

4 4

6 6

1

1

+ =

+ =Bi 5. Gii cc h phng trnh sau:

a)x xy y x y

x xy y x y

2 2

2 2 2

3( )

7( )

+ =

+ + =

b)xy x y x y

x y y x x y

2 22

2 1 2 2

+ + =

=

c)x y x y

x y x y

2 2

2 2

( )( ) 13

( )( ) 25

+ =

+ =d)

x x y x y

x y x xy

4 3 2 2

3 2

1

1

+ =

+ =

e)

x y x y xy xy

x y xy x

2 3 2

4 2

5

4

5(1 2 )

4

+ + + + =

+ + + =

f)x x y x y x

x xy x

4 3 2 2

2

2 2 9

2 6 6

+ + = +

+ = +

g)x x x y

x x y

2

2

( 2 )(3 ) 18

5 9 0

+ + =

+ + =h)

x y x y

x y x y

2 2

2 2

( )( ) 3

( )( ) 15

=

+ + =Bi 6. Tm m cc h phng trnh sau c nghim:

a)xy x y m

x y x y2 2( 1)( 1)

8

+ + =

+ + + =b)

x xy y

x xy y m

2 2

2 2

1

3 2

+ =

+ =

c)x y

x y y x x y m

1 1 3

1 1 1 1

+ + + =

+ + + + + + + =d) x y m

x y m

1 2

3

+ + =

+ =

e)

+=

=

mxyx

yxy

26

12

2

2

f)

x yx y

x y mx y

3 3

3 3

1 15

1 1 15 10

+ + + =

+ + + =

Bi 7. Tm m cc h phng trnh sau c nghim duy nht:

a)y m x

x m y

2

2

( 1)

( 1)

+ = +

+ = +b)

mx y

y

my x

x

22

22

2

2

= +

= +

c)x y m

x xy

2 0

1

=

+ =

d)y x x mx

x y y my

2 3 2

2 3 2

4

4

= +

= +

e)

=+

=+

)1(

)1(

2

2

xayxy

yaxxy (K cn v ) f)

+=+

+=+

axy

ayx2

2

)1(

)1( (K cn v )


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