(x/x+y-x/y-x-2xy/x.x-y.y)/(1/x-1/y) 分式化简

(x/x+y-x/y-x-2xy/x.x-y.y)/(1/x-1/y)
={[(x^2-xy)+(x^2+xy)-2xy]/(x^2-y^2)}/[(y-x)/xy]
={[2x^2-2xy]/(x-y)(x+y)}/[(y-x)/xy]
=2x/(x+y)/[(y-x)/xy]
=2x^2y/(y^2-x^2)

这个？？你加油啊，你一定行。

(x/x+y-x/y-x-2xy/x.x-y.y)/(1/x-1/y)
=((x*(x-y)-x*(x+y)-2xy)/(x^2-y^2))/(1/x-1/y)
=(-4xy/((x+y)(x-y)))/((y-x)/(xy))
=4x^2y^2/((x+y)(x-y)^2)