原式=2xy/(x+y)(x-y)+x(x-y)/(x+y)(x-y)
=(2xy+x²-xy)/(x+y)(x-y)
=(x²+xy)/(x+y)(x-y)
=x(x+y)/(x+y)(x-y)
=x/(x-y)3-x/x-2/{x+2-（x-2分之5）}其中x=2根号2采纳我，重新问