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3个回答
展开全部
2x²y-[x²y+2(x²y-xy)]+y
=2x²y-(x²y+2x²y-2xy)+y
=2x²y-(3x²y-2xy)+y
=2x²y-3x²y+2xy+y
=-x²y+2xy+y
=-y(x²-2x-1)
=-y[x²-(2x+1)]
=-(2011/2012)(x²-x²)
=0
=2x²y-(x²y+2x²y-2xy)+y
=2x²y-(3x²y-2xy)+y
=2x²y-3x²y+2xy+y
=-x²y+2xy+y
=-y(x²-2x-1)
=-y[x²-(2x+1)]
=-(2011/2012)(x²-x²)
=0
追问
确定对吗
追答
确定
展开全部
2x²y-[x²y+2(x²y-xy)]+y
=2x²y-[x²y+2xy(x-1)]+y
=2x²y-[x²y+2xy*2x]+y
=-3x²y+y
后面不会了
=2x²y-[x²y+2xy(x-1)]+y
=2x²y-[x²y+2xy*2x]+y
=-3x²y+y
后面不会了
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展开全部
原式=2x²y-[x²y+2xy(x-1)]+y=2x²y-[xy(x+2x-2)]+y=-x²y+2xy+y=y(1+2x-x²)=0
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