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已知:(x+2)²+|x+y-10|=0,求3x²y-[5x²y-﹙2xy-x²

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已知:(x+2)²+|x+y-10|=0,求3x²y-[5x²y-﹙2xy-x²y﹚-2x²]-xy的值
(x+2)²+|x+y-10|=0
x+2=0
x+y-10=0
x=-2
y=12
3x²y-[5x²y-﹙2xy-x²y﹚-2x²]-xy
=3x²y-(5x²y-2xy+x²y-2x²)-xy
=3x²y-(6x²y-2xy-2x²)-xy
=3x²y-6x²y+2xy+2x²+xy
=-x²y+3xy+2x²
=-4*12+3*(-2)*12+2*4
=-48-72+8
=-102