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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
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|x-1|+(y-1)²=0
x-1=0 y-1=0
x=1 y=1
(x²+4xy-2y²)-(x²+y)-2(y²+xy)-½
=(1+4-2)-(1+1)-2(1+1)-1/2
=3-2-4-1/2
=-3-1/2
=-7/2
x-1=0 y-1=0
x=1 y=1
(x²+4xy-2y²)-(x²+y)-2(y²+xy)-½
=(1+4-2)-(1+1)-2(1+1)-1/2
=3-2-4-1/2
=-3-1/2
=-7/2
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