数学：先化简再求值(x²+xy-2y²/x²-2xy+y²)-(1-xy²)(1+xy²)-x²÷1/y^4,其中x=根号3+根号2y=根号3-根号2

数学：先化简再求值(x²+xy-2y²/x²-2xy+y²)-(1-xy²)(1+xy²)-x²÷1/y^4,其中x=根号3+根号2y=根号3-根号2
数学：先化简再求值(x²+xy-2y²/x²-2xy+y²)-(1-xy²)(1+xy²)-x²÷1/y^4,其中x=根号3+根号2
y=根号3-根号2

数学：先化简再求值(x²+xy-2y²/x²-2xy+y²)-(1-xy²)(1+xy²)-x²÷1/y^4,其中x=根号3+根号2y=根号3-根号2
(x²+xy-2y²/x²-2xy+y²)-(1-xy²)(1+xy²)-x²÷1/y^4
=【（x+2y)(x-y)/(x-y)² 】-（1-x² y^4)-x² y^4
=(x+2y)/(x-y)-1+x² y^4-x² y^4
=(x+2y)/(x-y)-1
=【(x+2y)-（x-y)】/(x-y)
=3y/(x-y)
你只告诉了x,没告诉y,自己在对照题目,将x.y带入求出就行

原题不是这样吧

.

请问这是等式吗？没有等式那就是无穷解

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