已知x²+4y²-4x+4y+5=0,求：(x^4-y^4)/[(2x-y)(x+y)]*(2x-y)/(xy-y²)/[(x²+y²)²/y²]

已知x²+4y²-4x+4y+5=0,求：(x^4-y^4)/[(2x-y)(x+y)]*(2x-y)/(xy-y²)/[(x²+y²)²/y²]
已知x²+4y²-4x+4y+5=0,求：
(x^4-y^4)/[(2x-y)(x+y)]*(2x-y)/(xy-y²)/[(x²+y²)²/y²]

已知x²+4y²-4x+4y+5=0,求：(x^4-y^4)/[(2x-y)(x+y)]*(2x-y)/(xy-y²)/[(x²+y²)²/y²]
x^2+4y^2-4x+4y+5=0.
(x-2)^2-4+4*(y+1/2)^2-1+5=0.
(x-2)^2+4(y+1/2)^2=0.
(x-2)^2=0,x=2；
(y+1/2)^2=0,y=-1/2.
原式=[(x-y)(x+y)((X^2+Y^2)]/[(2x-y)(x+y)]*(2x-y)/y(x-y)/[x^2+y^2)^2/y^2].分子、分母约去公因子后,得：
原式=y/(x^2+y^2).(1)
将x=2,y=-1/2代人(1),得：
原式=(-1/2)/[(2^2+(-1/2)^2].
=--2/17.
∴(x^4-y^4)/[(2x-y)(x+y)]*(2x-y)/(xy-y^2)/[(x^2+y^2)^2/y^2]=-2/17 .

从第一个式子可以得到x=2，y=-1/2；然后下面的式子，你自己算吧

等式x²+4y²-4x+4y+5=0可化为：
x²-4x+4+4y²+4y+1=0
即(x-2)²+(2y+1)²=0
则有：x-2=0，2y+1=0
解得：x=2，y=-1/2
所以：
(x^4-y^4)/[(2x-y)(x+y)]*(2x-y)/(xy-y²)/[(x²...

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等式x²+4y²-4x+4y+5=0可化为：
x²-4x+4+4y²+4y+1=0
即(x-2)²+(2y+1)²=0
则有：x-2=0，2y+1=0
解得：x=2，y=-1/2
所以：
(x^4-y^4)/[(2x-y)(x+y)]*(2x-y)/(xy-y²)/[(x²+y²)²/y²]
=(x²+y²)(x²-y²)/[(2x-y)(x+y)]*(2x-y)/[y(x-y)] *y²/(x²+y²)²
=y/(x²+y²)
=(-1/2)÷(4 + 1/4)
=(-1/2)÷(17/4)
=(-1/2)×(4/17)
=-2/17

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