计算.(x^2-y^2/2x^2+4xy+2y^2)*(2x+2y/x^2-xy)^2
问题描述：
计算.(x^2-y^2/2x^2+4xy+2y^2)*(2x+2y/x^2-xy)^2
答
(x^2-y^2)/(2x^2+4xy+2y^2)*(2x+2y)/(x^2-xy)^2
=(x^2-y^2)/[2(x^2+2xy+y^2)]*2(x+y)/[x^2(x-y)^2 ]
=(x-y)(x+y)/[2(x+y)]^2*2(x+y)/[x^2(x-y)^2 ]
=(x-y)(x+y)/(x+y)*1/[x^2(x-y)^2 ]
=(x-y)/[x^2(x-y)^2 ]
=1/[x^2(x-y)]你题目看错了。后面是2x+2y/x^2-xy这个分数的平方、不是x^2-xy的平方。(x^2-y^2)/(2x^2+4xy+2y^2)*[(2x+2y)/(x^2-xy)]^2 =(x^2-y^2)/[2(x^2+2xy+y^2)]*[2(x+y)/(x^2-xy)]^2=(x^2-y^2)/[2(x+y)^2]*4(x+y)^2/(x^2-xy)]^2=(x^2-y^2)*2/(x^2-xy)^2=2(x-y)(x+y)/(x^2-xy)^2=2(x-y)(x+y)/[x^2(x-y)^2]=2(x+y)/[x^2(x-y)]