已知|x-1|+(y-1)²=0求代数式(x²+4xy-2y²)-(x²+y)-2(y²+xy)-½(x-8y²)的值
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已知|x-1|+(y-1)²=0求代数式(x²+4xy-2y²)-(x²+y)-2(y²+xy)-½(x-8y²)的值
已知|x-1|+(y-1)²=0求代数式(x²+4xy-2y²)-(x²+y)-2(y²+xy)-½(x-8y²)的值
已知|x-1|+(y-1)²=0求代数式(x²+4xy-2y²)-(x²+y)-2(y²+xy)-½(x-8y²)的值
因为|x-1|+(y-1)²=0,且|x-1|>=0,(y-1)²>=0.
所以x=1,y=1
将x=1,y=1代入原代数式(x²+4xy-2y²)-(x²+y)-2(y²+xy)-½(x-8y²)得
(1²+4×1×1-2×1²)-(1²+1)-2(1²+1×1)-½(1-8×1²)
=(1+4-2)-(1+1)-2(1+1)-½(1-8)
=3-2-4+7/2
= -3+7/2
=1/2