已知x^2+y^2-2x+4y+5=0,求(x^4-y^4\2x^2+xy-y^2)乘(2x-y/xy-y^2)除以（x

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已知x^2+y^2-2x+4y+5=0,求(x^4-y^4\2x^2+xy-y^2)乘(2x-y/xy-y^2)除以（x^2+y^2\y)^2的值.
麻烦您啦,我不会打那些个符号,所以您费心了...

$已知x^2+y^2-2x+4y+5=0,求(x^4-y^4\2x^2+xy-y^2)乘(2x-y/xy-y^2)除以（x$

x^2+y^2-2x+4y+5 =x^2-2x+1+y^2+4y+4 =(x-1)^2+(y+2)^2 =0
因为任意数的平方都是大于或等于0的,所以x-1=0并且y+2=0才能满足上述等式.
所以,x=1,y=-2.
(x^4-y^4\2x^2+xy-y^2)乘(2x-y/xy-y^2)除以（x^2+y^2\y)^2把它分解因式,分子分母约分,得
=y/(x-y)(x²+y²)
把 x=1,y=-2代入上式得-2/15

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