x-y分之1-x+y分之1)除以x的平方-y的平方分之xy的平方顺便再给我解释下(x-y分之1-x+y分之1)除以x^2-2xy+y^2分之2y.最后答案为2y/(x-y)(x+y)×(x-y)^2/2y=x-y/x+y为什么(x-y分之1-x+y分之1)=2y/(x-y)(x+y)
问题描述：
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
顺便再给我解释下(x-y分之1-x+y分之1)除以x^2-2xy+y^2分之2y.
最后答案为2y/(x-y)(x+y)×(x-y)^2/2y=x-y/x+y
为什么(x-y分之1-x+y分之1)=2y/(x-y)(x+y)
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
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)