已知x^2+y^2-2x+2y=-2,求代数式x^2007+y^2008的值

方程移项配方得:(x-1)^2+(y+1)^2=0,两非负数之和是零,则这两数均是零,得x=1,y=-1,所以x^2007+y^2008=1^2007+(-1)^2008=2

x^2+y^2-2x+2y=-2
x^2+y^2-2x+2y+2=0
(x^2-2x+1)+(y^2+2y+1)=0
(x-1)^2+(y+1)^2=0
x-1=0,y+1=0
x=1,y=-1

x^2007+y^2008
=1^2007+(-1)^2008
=1+1
=2

我做过的 得401500

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