xy/x^2-y^2÷(1/x-y-1/x+y)要过程
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是不是x^2-y^2有个括号，如有，解如下：
原式=xy/(x²-y²)÷[(1/(x-y)-1/(x+y)]
=xy/(x²-y²)÷[x+y-(x-y)]/(x²-y²)
=xy/(x²-y²)×(x²-y²)/(2y)
=x/2
原式=xy/(x²-y²)÷[(1/(x-y)-1/(x+y)]
=xy/(x²-y²)÷[x+y-(x-y)]/(x²-y²)
=xy/(x²-y²)×(x²-y²)/(2y)
=x/2
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