有理数x,y满足2x^2-2xy+y^2+2x+1=0则(xy)^2004的值为A 1 B 0C -1D 2

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有理数x,y满足2x^2-2xy+y^2+2x+1=0则(xy)^2004的值为A1B0C-1D2有理数x,y满足2x^2-2xy+y^2+2x+1=0则(xy)^2004的值为A1B0C-1D2有理

有理数x,y满足2x^2-2xy+y^2+2x+1=0则(xy)^2004的值为A 1 B 0C -1D 2
有理数x,y满足2x^2-2xy+y^2+2x+1=0则(xy)^2004的值为
A 1
B 0
C -1
D 2

有理数x,y满足2x^2-2xy+y^2+2x+1=0则(xy)^2004的值为A 1 B 0C -1D 2
2x^2-2xy+y^2+2x+1=0
(x^2-2xy+y^2)+(x^2+2x+1) =0
得(x-y)^2+(x+1)^2=0
所以x-y=0
x+1=0
那么x=-1,y=-1
原式=1^2004=1
故选a

2x^2-2xy+y^2+2x+1
=x^2-2xy+y^2+x^2+2x+1
=(x-y)^2+(x+1)^2
=0
所以(x-y)=0,(x+1)=0,可得x=-1,y=-1,所以(xy)^2004=1^2004=1,所以选a

a

(X-Y)^2+(x+1)^2=0
x=-1
Y=-1
(xy)^2004=1