已知X2-2x+y2+6y+10=0,求（x2-2xy）/(xy+y2)的值
问题描述：
已知X2-2x+y2+6y+10=0,求（x2-2xy）/(xy+y2)的值
答
X2-2x+y2+6y+10=0
x²-2x+1+y²+6y+9=0
(x-1)²+(y+3)²=0
x-1=0
y+3=0
∴x=1
y=-3
（x2-2xy）/(xy+y2)
=(1+6)/(-3+9)
=7/6
答
x²-2x+y²+6y+10=0,变换得(x-1)²+(y+3)²=0,∴x=1,y=-3
∴
（x2-2xy）/(xy+y2)
=(1²-2*(-3))/(1*(-3)+(-3)²)
=7/6