Question

求解方程组，高分悬赏！

1. 2x^2-4xy+y^2+2x-y+2=0
x^2-2xy-y^2+x-2y+4=0

2. xy-y^2+x^2=1
x^2+y^2+4xy=2

Answer 1

1， ①式-2*②式得
3y²+3y-6=0 => y²+y-2=0 => y=1或-2
y=1时，带入②式得
x²-2x-1+x-2+4=0 => x²-x+1=0 => x无实根
y=-2时，带入②式得
x²+4x-4+x+4+4=0 => x²+5x+4=0 => x=-1或-2
∴解为(x,y)=(-1,-2)或(-2,-2)
2，2*①式-②式得
x²-2xy-3y²=0 => (x-3y)(x+y)=0 => x=3y或x=-y
x=3y带入①式得
3y²-y²+9y²=1 => 11y²=1 => y=±1/√11，x=±3/√11
x=-y带入①式得
-y²-y²+y²=1 => -y²=1 => y²=-1 => y无实根
∴解为(x,y)=(3/√11,1/√11)或(-3/√11,-1/√11)

Answer 2

1.  2x^2-4xy+y^2+2x-y+2=0  (1)
       x^2-2xy-y^2+x-2y+4=0  (2)

   (2)*2得：2x^2-4xy-2y^2+2x-4y+8=0(3)

   (1)-(3)得：3y^2+3y-6=0，所以y^2+y-2=0，所以y=-2或y=1，代入(2)可求得对应的x

   
   

Answer 3

1、1式-2*2式
3y*2+3y-6=0 y=-2或y=1
再带进去解x
2、2*1式-2式
x^2-2xy-3y^2=0
(x-3y)(x+y)=0
x=3y或x=-y
再带回去解