giai baitap pt hpt

8/6/2019 Giai Baitap PT HPT

1/9

www.ybfx.wordpress.comTI LIU LTH 2011 CHUYN : PHNG TRNH - H PHNG TRNH

(Di y ch l cch gii gi , cha phi l cch ngn gn nht. Mi kin ng gp xin cc bn [email protected].)

Phn I: Gii cc h phng trnh sau:

1. 3(x y) = 2xy (1)

2x y2 = 8 (2)

Gii:iu kin: xy 0; x y(1) 3(x y)2 = 4xy (3x y)(x 3y) = 0 x = 3y hay x = y

3

Thay vo (2) H c nghim:

x = 6y = 2

hay

x = 12y = 4

2.

x2

+ y2

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

Gii:

T h y = 0

x2 + 1

y+ x + y = 4

(x + y)2 2 x2 + 1

y= 7

t

u =x2 + 1

y

v = x + y

u + v = 4

v2 2u = 7

u = 1

v = 3hay

u = 9

v = 5

x = 1

y = 2hay

x = 2y = 5

3. 2log1x(xy 2x + y + 2) + log2+y(x

2

2x + 1) = 6

log1x(y + 5) 2log2+y(x + 4) = 1

Gii:

iu kin:

xy 2x + y + 2 > 0x2 2x + 1 > 0y + 5 > 0x + 4 > 00 < 1 x = 10 < 2 + y = 1

(*)

1


2/9

H cho

log1x (2 + y) + log2+y (1 x) 2 = 0 (1)log1x(y + 5) 2log2+y(x + 4) = 1 (2)

t: t = log2+y (1 x) th (1) t = 1 y = x 1

Thay vo (2) ta c:

x = 0

y = 1(khng tha (*))

hay

x = 2y = 1

(tha (*))

4.

x3y(1 + y) + x2y2(2 + y) + xy3 30 = 0x2y + x(1 + y + y2) + y 11 = 0

Gii:

H cho

xy(x + y)2 + x2y2(x + y) = 30

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

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

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

t:

u = x + y

v = xy

uv(u + v) = 30

uv + u + v = 11

uv = 5 u + v = 6uv = 6 u + v = 5

Vi uv = 5 u + v = 6 x = 5

21

2 y = 5 +

21

2hay x =

5 +

21

2 y = 5

21

2

Vi uv = 6 u + v = 5 x = 1 y = 2 hay x = 2 y = 1

5. x3 + 4y = y3 + 16x (1)

1 + y2 = 5(1 + x2) (2)

Gii:T (2) y2 5x2 = 4 (3). Thay vo (1)

x = 0x2 5xy 16 = 0

Vi x = 0 y = 2Vi x2 5xy 16 = 0 y = x

2 165x

, thay vo (3) ta c:

124x4 + 132x2

256 = 0

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

6.

log2

x + y = 3log8(

x y + 2)

x2 + y2 + 1

x2 y2 = 3

Gii:iu kin: x + y > 0; x y 0

H cho

x + y =

x y + 2

x2 + y2 + 1x2 y2 = 32


3/9

t:

u = x + yv = x y . Ta c h:

uv = 2u2 + v2 + 2

2uv = 3

u + v = 2

uv + 4

u > v(u + v)2 2uv + 2

2uv = 3

uv + 8

uv + 9uv = 3 uv = 0 u + v = 4 u = 4 v = 0 x = y = 2

7.

1 + x3y3 = 19x3 (1)

y + xy2 = 6x2 (2)

Gii:V x = 0 y = 0 khng tha (2) nn ta ly (1) chia cho (2) v theo v, ta c:

(1 + xy)(1 xy + x2y2)y(1 + xy)

= 196

x

6x2y2 + 13xy + 6 = 0 xy =

3

2

(1)

x = 1

2 y = 3xy = 2

3

(1) x = 13 y = 2

8.

4y

x+xy = 32

log3 (x y) = 11

2log3(x + y)

Gii:

iu kin: xy = 0x y

H cho 2

x

y+

y

x

= 5

x2 y2 = 3

t: t =x

y 2t2 5t + 2 = 0 t = 1

2 t = 2 x = 2 y = 1.

9. 2x2y + y3 = 2x4 + x6 (1)

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

Gii:

(1) 2x2(y x2) + (y x2)(y2 + yx2 + x4) = 0

y = x2

2x2 + y2 + yx2 + x4 = 0

3


4/9

y = x2y +

x2

2

2+

3x4

4+ 2x2 = 0 x = y = 0 (loi v khng tha (2))

Vi y = x2, thay vo (2) ta c:

(x + 2)

x2 + 1 = x2 + 2x + 1 (xx2 + 1 2x) + [2x2 + 1 (x2 + 1)] = 0

(x2 + 1 2)(xx2 + 1) = 0 x = 3 y = 3.

10.

x2 + 2 +

y2 + 3 + x + y = 5

x2 + 2 +

y2 + 3 x y = 2

Gii:

Cng v tr tng v 2 phng trnh ca h, ta c h tng ng:

x2 + 2 +

y2 + 3 =

7

2

x + y =3

2

x2 + 2 +

3

2 x

2+ 3 =

7

2

y =3

2 x

...

x =

1

2 y = 1

x =17

20 y = 13

20

11.

x3 6x2y + 9xy2 4y3 = 0 (1)

x y +x + y = 2 (2)

Gii:V y = 0 khng tha h nn ta chia 2 v ca (1) cho y3:

(1)

x

y

3 6

x

y

2+ 9

x

y 4 = 0

x

y 1

2x

y 4

= 0

x = y x = 4y

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

Vi x = 4y (2) x = 32 815 y = 8 215

12.

lg2 x = lg2 y + lg2 (xy)

lg2 (x y) + lg x. lg y = 0

Gii:iu kin: x > 0 y > 0

4


5/9

H cho

lg2 x = lg2 y + (lg x + lg y)2

lg2 (x y) + lg x. lg y = 0

lg y (lg x + lg y)2

lg2 (x y) + lg x. lg y = 0

lg y = 0

lg2 (x y) = 0

lg x + lg y = 0

lg2 (x y) + lg x. lg y = 0

x = 2y = 1

x =

2

y =

2

2

13.

x2 + 5x + y = 9

3x3 + x2y + 2xy + 6x2 = 18

Gii:

H cho y = 9 x2

5x

x4 + 4x3 5x2 18x + 18 = 0

x = 1 y = 3x =

3

y = 15

x = 17 y = 6 + 37x = 1 +7 y = 6 37

14.

9x2 4y2 = 5log5 (3x + 2y) log3 (3x 2y) = 1

Gii:

H cho

log5 (3x + 2y) + log5 (3x 2y) = 1

log5 (3x + 2y) log3 5. log5 (3x 2y) = 1 log5 (3x + 2y) = 1

log5 (3x 2y) = 0 x = 1

y = 1

15.

x 2y xy = 0

x 1 +4y 1 = 2

Gii:iu kin: x 1 y 1

4

H cho

(

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

x 1 +4y 1 = 2 (

x +

y)(

x 2y) = 0

x 1 + 4y 1 = 2

x 2y = 0

x 1 + 4y 1 = 2 x = 2

y =1

2

16.

8x3y3 + 27 = 7y3 (1)

4x2y + 6x = y2 (2)

5


6/9

Gii:T (1) y = 0.(2) 4x2y2 + 6xy = y3

t: t = xy, t h cho 8t3 + 27 = 4t2 + 6

t = 32

t =1

2

t =9

2

Vi t = 32 y = 0 (loi)

Vi t =1

2 x = 1

2 3

4 y = 34

Vi t =9

2 x = 3

2 3

4 y = 3 34

17. x

2

+ y2

+

2xy

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

Gii:iu kin: x + y > 0.

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

1 1x + y

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

y = 1

x, thay vo (2)

x = 1 y = 0

x = 2 y = 3

18.

log20112yx

= x 2y (1)

x3 + y3

xy= x2 + y2 (2)

Gii:iu kin: xy > 0.

T (2) ta c: x3 + y3 = xy(x2 + y2) > 0 x > 0 y > 0

(1) 2yx

= 2011x2y x.2011x = 2y.20112y (*)

Xt hm s: f(t) = t.2011t; (t > 0). Ta c: f(t) = 2011t

1 +t

ln 2011

> 0, t > 0

f(t) ng bin khi t > 0. Do t () f(x) = f(2y) x = 2y

Thay vo (2) ta c: y

5y 9

2

= 0 y = 9

10 x = 9

5

6


7/9

19.

log2 (x

2 + y2) = 1 + log2 (xy)

3x2xy+y2 = 81

Gii:iu kin: xy > 0.

H cho x2 + y2 = 2xy

x

2

xy + y2

= 4

x = y

x

2

= 4

x = y = 2

x = y = 2(v xy > 0)

20.

2y2 x2 = 12x3 y3 = 2y x

Gii:T h cho 2x3 y3 = 1(2y x) = (2y2 x2)(2y x) x3 + 2x2y + 2xy2 5y3 = 0 ()Nhn thy: y = 0 khng tha h, chia 2 v ca () cho y3 ta c:

xy3

+ 2x

y2

+ 2x

y 5 = 0 xy = 1

2y2 x2 = 1x

y= 1

x = y = 1 x = y = 1

21.

3(x3 y3) = 4xyx2y2 = 9

Gii:x2y2 = 9 xy = 3.Khi xy = 3, ta c: x3 + (y3) = 4 v x3.(y3) = 27 x3,y3 l cc nghim ca phng trnh: X2 4X 27 = 0 X = 231 x = 3

2 +

31 y = 3

231 hoc x = 3

231 y = 3

2 +

31

Khi xy = 3, ta c: x3 + (y3) = 4 v x3.(y3) = 27 x3,y3 l cc nghim ca phng trnh: X2 + 4X + 27 = 0. Phng trnh ny v nghim.

22. y2 + x = x2 + y (1)

2x = 3y+1 (2)

Gii:

Ta c: (1) (y x)(y + x 1) = 0

x = y

y = 1 xVi y = x, thay vo h x = y = log 2

3

3

Vi y = 1 x, thay vo h x = log6 9 y = 1 log6 9

7


8/9

23.

3

x2 + y2 1 +2y

x= 1

x2 + y2 +4x

y= 22

Gii:iu kin: xy = 0 x2 + y2 1 = 0.

t: u = x2

+ y2

1; v =x

y , thay vo h ta c:

3

u+

2

v= 1

u + 4v = 21 2v2 13v + 21 = 0

v = 3

v =7

2

Vi v = 3 u = 9 ta gii c

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

Vi v =7

2 u = 7 ta gii c

x = 142

53 y = 42

53

x = 14

2

53 y = 4

2

53

24.

23x+1 + 2y2 = 3.2y+3x (1)

3x2 + xy + 1 =

x + 1 (2)

Gii:

Ta c: (2)

x + 1 0

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

Vi x = 0, thay vo (1) y = log28

11

Vi

x 1x = 0 y = 1 3x

, thay vo (1) 23x+1 + 23x1 = 6 23x+1 = 3 + 22

(v x 1 23x+1 14

)

T ta tm c x =1

3

log2 (3 + 2

2) 1 y = 2 log2 (3 + 22)

25.

log4 (x

2 + y2) log4 (2x) + 1 = log4 (x + 3y)log4 (xy + 1) log4 (4y2 + 2y 2x + 4) = log4 (xy ) 1

Gii:

8


9/9

iu kin:

x > 0;

x

y> 0

4y2 + 2y 2x + 4 > 0

H cho

4(x2 + y2)

2x= x + 3y (1)

xy + 1

4y2 + 2y

2x + 4

=x

4y(2)

V x > 0 nn (1) x2 + 2y2 3xy = 0

x

y

2 3

x

y

+ 2 = 0 x

y= 1 x

y=

1

2

Vix

y= 1, thay vo (2) ta tm c x = y = , l s thc dng bt k.

Vix

y=

1

2, thay vo (2) ta tm c x = 2 y = 1.

9

giai baitap pt hpt

Documents