有这样一道题：“计算（2x^3-3x^2y-2xy^2)-(x^3-2xy^2+y^3)+(-x^3+3x^2y-y^3)的值，

显示全部楼层 · 2011-7-21 19:30:30

有这样一道题："计算（2x^3-3x^2y-2xy^2)-(x^3-2xy^2+y^3)+(-x^3+3x^2y-y^3)的值，其中x=1/2,y=-1"甲同学把“x=1/2”错抄成“x=-1/2”但他计算的结果也是正确的。试说明理由，并求出结果。

千问 · 2011-7-21 19:30:30

将（2x^3-3x^2y-2xy^2)-(x^3-2xy^2+y^3)+(-x^3+3x^2y-y^3)化简会发现：（2x^3-3x^2y-2xy^2)-(x^3-2xy^2+y^3)+(-x^3+3x^2y-y^3)=-2y^3也就是说其值与x无关 sox抄错结果也正确...

千问 · 2011-7-21 19:30:30

将（2x^3-3x^2y-2xy^2)-(x^3-2xy^2+y^3)+(-x^3+3x^2y-y^3)化简会发现：（2x^3-3x^2y-2xy^2)-(x^3-2xy^2+y^3)+(-x^3+3x^2y-y^3)=-2y^3也就是说其值与x无关 sox抄错结果也正确...

千问 · 2011-7-21 19:30:30

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