已知x^2+y^2-2x+4y+5=0.求{(x^4-y^4)/[(x+y)(2x-y)]}*{[(2x-y)/(xy-

由题得：(x-1)²+(y+2)²=0
所以,x=1,y=-2
{(x^4-y^4)/[(x+y)(2x-y)]}*{[(2x-y)/(xy-y^2)]}/{[x^2+y^2]/y}^2={[(x-y)(x+y)(x²+y²)]/[(x+y)(2x-y)]}*[(2x-y)/(x-y)y]*[y²/(x²+y²)²]=y/(x²+y²)=(-2)/(1+4)=-2/5

x^2+y^2-2x+4y+5=0
(x-1)^2+(y+2)^2=0
满足上式的唯一条件是
x-1=0 x=1
y+2=0 y=-2
代入{(x^4-y^4)/[(x+y)(2x-y)]}*{[(2x-y)/(xy-y^2)]}/{[x^2+y^2]/y}^2
={(x^2+y^2)(x+y)(x-y)/[(x+y)(2x-y)]}*{y(2x-y)(x-y)]/[y^2(x^2+y^2)^2]
=(x-y)^2/[y(x^2+y^2)]
=(1+2)^2/[(-2)(1+4)]
=9/(-10)
=-9/10

x^2+y^2-2x+4y+5=0
(x-1)²+ (y-2)²=0
x-1=0
y-2=0
x=1
y=2
自己代入一下算出来就是

由题得：(x-1)²+(y+2)²=0
所以，x=1, y=-2
{(x^4-y^4)/[(x+y)(2x-y)]}*{[(2x-y)/(xy-y^2)]}/{[x^2+y^2]/y}^2={[(x-y)(x+y)(x²+y²)]/[(x+y)(2x-y)]}*[(2x-y)/(x-y)y]*[y²/(x²+y²)²]=y/(x²+y²)=(-2)/(1+4)=-2/5