原式=[(m-n)/(m-n)²-n(m+n)/(m+n)(m-n)]·mn/(n-1)
=[1/(m-n)-n/(m-n)]·mn/(n-1)
=(1-n)/(m-n)·mn/[-(1-n)]
=-mn/(m-n)
初二数学问题,计算题
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原式=[(m-n)/(m-n)²-n(m+n)/(m+n)(m-n)]·mn/(n-1)
=[1/(m-n)-n/(m-n)]·mn/(n-1)
=(1-n)/(m-n)·mn/[-(1-n)]
=-mn/(m-n)