已知|x-1|+(y-1)²=0求代数式(x²+4xy-2y²)-(x²+y)-
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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)-](/uploads/image/z/1199749-13-9.jpg?t=%E5%B7%B2%E7%9F%A5%7Cx-1%7C%2B%EF%BC%88y-1%EF%BC%89%26%23178%3B%3D0%E6%B1%82%E4%BB%A3%E6%95%B0%E5%BC%8F%EF%BC%88x%26%23178%3B%2B4xy-2y%26%23178%3B%EF%BC%89-%EF%BC%88x%26%23178%3B%2By%EF%BC%89-)
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
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
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