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  • x-y分之1-x+y分之1)除以x的平方-y的平方分之xy的平方

    x-y分之1-x+y分之1)除以x的平方-y的平方分之xy的平方
    順便再給我解釋下(x-y分之1-x+y分之1)除以x^2-2xy+y^2分之2y.
    最后答案為2y/(x-y)(x+y)×(x-y)^2/2y=x-y/x+y
    為什么(x-y分之1-x+y分之1)=2y/(x-y)(x+y)
    數(shù)學(xué)人氣:644 ℃時(shí)間:2020-05-05 11:48:27
    優(yōu)質(zhì)解答
    [1/(x-y)-1/(x+y)]/[xy^2/(x^2-y^2)]
    =[(x+y-x+y)/(x-y)(x+y)]/[xy^2/(x-y)(x+y)]
    =[2y/(x-y)(x+y)]/[xy^2/(x-y)(x+y)]
    =2y/(x-y)(x+y)*(x-y)(x+y)/xy^2
    =2y/xy^2
    =2/xy
    [1/(x-y)-1/(x+y)]/[2y/(x^2-2xy+y^2)]
    =[(x+y-x+y)/(x-y)(x+y)]/[2y/(x-y)^2]
    =2y/(x-y)(x+y)*(x-y)^2/2y
    =(x-y)^2/(x-y)(x+y)
    =(x-y)/(x+y)
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