问题描述: 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 1个回答 分类:数学 2014-10-26 问题解答: 我来补答 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 展开全文阅读