已知(x-1)²+|y+2|=0,求代数式(x-½y-1)(x-½y+1)-(x-½y-1)²的值.求详细步骤!
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已知(x-1)²+|y+2|=0,求代数式(x-½y-1)(x-½y+1)-(x-½y-1)²的值.求详细步骤!
已知(x-1)²+|y+2|=0,求代数式(x-½y-1)(x-½y+1)-(x-½y-1)²的值.
求详细步骤!
已知(x-1)²+|y+2|=0,求代数式(x-½y-1)(x-½y+1)-(x-½y-1)²的值.求详细步骤!
(x-1)的平方+|y+2|=0 → x=1, y=-2
∴(x-1/2y-1)(x-1/2y+1)-(x-1/2y-1)2=1
因为(x-1)²+|y+2|=0
所以x-1=0 y+2=0
则x=1 y=-2
故(x-½y-1)(x-½y+1)-(x-½y-1)²=2
因为(x-1)²+|y+2|=0,
所以x-1=0 y+2=0
所以x=1,y=-2
将x=1,y=2带入代数式中,
原式=1x3-1=3-1=2