3 2X Y X XY 1− + = (I) 2 2 3( X XY) X Y 1(I) ⇔  − +  + = − + + =..

2. Giải hệ:

3 2

x y x xy 1

− + =

 (I)

2 2 3

( x xy) x y 1

(I) ⇔  − +  + =

 − + + =

2 3



Đặt u = − x

2

+ xy, v = x

3

y

v u 1 u 0 u 1

2

u v 1

(I) thành      + = + = ⇔      = − + − = ⇔    = = ∨    = =

v 1 v 0

u v 1 u u 0

Do đĩ hệ đã cho tương đương:

2 2

 − + =  − + =  =  =

y x y 0

x xy 0 x xy 1

 ∨  ⇔  ∨ 

   

4 2

3 3

x 1 x 1(vn)

= = −

 

x y 1 x y 0

= =

   

 

x 1 x 1

 

⇔   = ∨   = −

y 1 y 1

Câu III:

a ( 1,2,0)