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