计算：(1/x-y-1/x+y)/2y/x^2-2xy+y^2,x=1+根号2,y=1-根号2

网友回答

[1/(x-y)-1/(x+y)]/[2y/(x^2-2xy+y^2)]
={[(x+y)-(x-y)]/[(x+y)(x-y)]}*(x^2-2xy+y^2)/2y
={2y/[(x+y)(x-y)]}*{(x-y)^2/2y}
=(x-y)/(x+y）
x=1+根号2,y=1-根号2,代入
原式=【(1+根号2）-（1-根号2)】/【(1+根号2）+（1-根号2)】
=(2根号2)/2
=根号2