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,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