求！速度！今天晚上就要有！谢谢了！已知xy=2,求(1/x-y+1/x+y)÷x²y/x²-y²的值。
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解：当xy=2时
原式=
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这个,题目是不是这样的[1/(x-y)+1/(x+y)]÷x²y/(x²-y²)
如果是的话
[1/(x-y)+1/(x+y)]÷x²y/(x²-y²)
=[2x/(x²-y²)]×(x²-y²)/(x²y)
=2/(xy)
=1
如果是的话
[1/(x-y)+1/(x+y)]÷x²y/(x²-y²)
=[2x/(x²-y²)]×(x²-y²)/(x²y)
=2/(xy)
=1
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(1/x-y+1/x+y)÷x²y/x²-y²
=(x+y+x-y)(x²-y²)/[x-y)(x+y)x²y]
=(2x)((x²-y²)/[(x²-y²)x²y]
=2/xy
=1
=(x+y+x-y)(x²-y²)/[x-y)(x+y)x²y]
=(2x)((x²-y²)/[(x²-y²)x²y]
=2/xy
=1
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[1/(x-y)+1/(x+y)]÷x²y/(x²-y²)
=(x+y+x-y)/(x²-y²)÷ x²y/(x²-y²)
=2x/(x²-y²)x(x²-y²)/ x²y
=2/xy
=1
=(x+y+x-y)/(x²-y²)÷ x²y/(x²-y²)
=2x/(x²-y²)x(x²-y²)/ x²y
=2/xy
=1
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