已知x^2+y^2-2x+2y=-2,求代数式x^2005+x^2006的值

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

所以x^2005+x^2006
=1+1
=2

x^2+y^2-2x+2y=-2
x^2-2x+y^2+2y=-2
(x^2-2x+1)-1+(y^2+2y+1)-1=-2
(x-1)^2+(y+1)^2=0
(x-1)^2=0 (y+1)^2=0
x=1 y=-1
所以 原式=x^2005+x^2006
1^2005+1^2006=2

x^2+y^2-2x+2y=-2
（x-1）^2+(y+1)^2=0
x=1 y=-1
结果显而易见