已知:x,y满足(x-1)*(x-1)+|y+1|=0.求代数值2(x*x-y*y+1)-2(*x*x+y*y)+xy的
2个回答
(x-1)^2+|y+1|=0 所以x-1=0 x=1,y+1=0 y=-1;
所以 原式=2(x*x-y*y-x*x-y*y)+2+xy
=-4(-1)*(-1)+2+1*(-1)
=5
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