题目

已知x²+y²-2x+4y+5=0,求X^4-y^4/2x^2+xy-y²·2x-y/xy-y^2÷(x^2+y^2)^2的值

提问时间：2020-10-17

答案

x²+y²-2x+4y+5=0
配方：(x-1)²+(y+2)²=0
所以只有x-1=0且y+2=0
∴x=1,y=-2
∴(X^4-y^4)/(2x^2+xy-y²)·[(2x-y)/(xy-y^2)]÷(x^2+y^2)^2
=(x²+y²)(x²-y²)/[(2x-y)(x+y)]× (2x-y)/[y(x-y)] ÷(x²+y²)²
=(x²-y²)/(x+y)× 1/[y(x-y)] ÷(x²+y²)
=(x+y)(x-y)/(x+y)×1/[y(x-y)]÷(x²+y²)
=1/y÷(x²+y²)
=1/[y(x²+y²)]
=1/[-2(1+4)]
=-1/10