化简m/(m+n）²+n/m²-n²-2n²/（m+n）（m²-n²﹚

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化简m/(m+n）²+n/m²-n²-2n²/（m+n）（m²-n²﹚
化简m/(m+n）²+n/m²-n²-2n²/（m+n）（m²-n²﹚

化简m/(m+n）²+n/m²-n²-2n²/（m+n）（m²-n²﹚
楼主所给式子,是m/(m+n)²+n/(m²-n²)-2n²/[(m+n)(m²-n²)]
如果是的话：
m/(m+n)²+n/(m²-n²)-2n²/[(m+n)(m²-n²)]
=m/(m+n)²+n/[(m-n)(m+n)]-2n²/[(m+n)²(m-n)]
=[m(m-n)+n(m+n)-2n²]/[(m+n)²(m-n)]
=(m²-mn+mn+n²-2n²)/[(m+n)²(m-n)]
=(m²-n²)/[(m+n)²(m-n)]
=(m-n)(m+n)/[(m+n)²(m-n)]
=1/(m+n)

m/(m+n）²+n/（m²-n²）-2n²/[(m+n)(m²-n²﹚]
=m(m-n)/[(m+n)²(m-n﹚]+n(m+n)/[(m+n)²(m-n﹚]-2n²/[(m+n)²(m-n﹚]
=(m²-mn+mn+n²-2n²)/[(m+n)²(m-n﹚]
=(m²-n²)/[(m+n)(m²-n²﹚]
=1/(m+n)

m/(m+n)²+n/(m²-n²)-2n²/(m+n)(m²-n²)
=[m(m²-n²)+n(m+n)²-2n²(m+n)]/[(m+n)²(m²-n²)]
=[m(m²-n²)+(nm+n²-2n²)(m+n...

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m/(m+n)²+n/(m²-n²)-2n²/(m+n)(m²-n²)
=[m(m²-n²)+n(m+n)²-2n²(m+n)]/[(m+n)²(m²-n²)]
=[m(m²-n²)+(nm+n²-2n²)(m+n)]/[(m+n)²(m²-n²)]
=[m(m²-n²)+(nm-n²)(m+n)]/[(m+n)²(m²-n²)]
=[m(m²-n²)+n(m-n)(m+n)]/[(m+n)²(m²-n²)]
=[m(m²-n²)+n(m²-n²)]/[(m+n)²(m²-n²)]
=[(m+n)(m²-n²)]/[(m+n)²(m²-n²)]
=1/(m+n)

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