∵x^2+y^2-2x+4y+5=0
配方：
∴(x-1)^2+(y+2)^2=0
那么x-1=0且y+2=0
所以x=1,y=-2
∴(x^4-y^4)/(2x^2+xy-y^2)* （2x-y)/(xy-y^2) ÷[(x^2+y^2)^2/y]
=(x^2+y^2)(x^2-y^2)/[(2x-y)(x+y)]*(2x-y)/[x(x-y)]*y/(x^2+y^2)^2
=（x+y)(x-y)/(x+y) *1/[x(x-y)]*y/(x^2+y^2)
= 1/x* y/(x^2+y^2)
=-2/5
后面的式子,你用的敘述方式,可能理解起來(lái)會(huì)有歧義