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有这样一道题:“计算2x*x*x-3x*x*y-2xy*y-x*x*x+2xy*y-y*y*y-x*x*x+3x*x*y-y*y*y的值x=1/2,y=-1
人气:183 ℃ 时间:2020-05-21 09:05:56
解答
2x^3-3x^2y-2xy^2-x^3+2xy^2-y^3-x^3+3x^2y-y^3
=(2x^3-x^3-x^3)+(3x^2y-3x^2y)+(2xy^2-2xy^2)-y^3-y^3
=-2y^3=-2(-1)^3=2
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