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x=2009 y=1 甲同学你、把x=2009看成x=-2009.但结果仍正确.计算出最后的值,说说怎么回事?
来源:学生作业帮助网 编辑:作业帮 时间:2024/07/12 08:02:44
![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x=2009 y=1 甲同学你、把x=2009看成x=-2009.但结果仍正确.计算出最后的值,说说怎么回事?](/uploads/image/z/1260583-7-3.jpg?t=%E6%9C%89%E8%BF%99%E6%A0%B7%E4%B8%80%E9%81%93%E9%A2%98%EF%BC%9A%E2%80%9C%E8%AE%A1%E7%AE%97%EF%BC%882x%5E3-3x%5E2y-2xy%5E2%29-%28x%5E3-2xy%5E2%2By%5E3%29%2B%28-x%5E3%2B3x%5E2y-y%5E3%29%E7%9A%84%E5%80%BCx%3D2009++++y%3D1+%E7%94%B2%E5%90%8C%E5%AD%A6%E4%BD%A0%E3%80%81%E6%8A%8Ax%3D2009%E7%9C%8B%E6%88%90x%3D-2009.%E4%BD%86%E7%BB%93%E6%9E%9C%E4%BB%8D%E6%AD%A3%E7%A1%AE.%E8%AE%A1%E7%AE%97%E5%87%BA%E6%9C%80%E5%90%8E%E7%9A%84%E5%80%BC%2C%E8%AF%B4%E8%AF%B4%E6%80%8E%E4%B9%88%E5%9B%9E%E4%BA%8B%3F)
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x=2009 y=1 甲同学你、把x=2009看成x=-2009.但结果仍正确.计算出最后的值,说说怎么回事?
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
x=2009 y=1 甲同学你、把x=2009看成x=-2009.但结果仍正确.计算出最后的值,说说怎么回事?
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x=2009 y=1 甲同学你、把x=2009看成x=-2009.但结果仍正确.计算出最后的值,说说怎么回事?
(2x^3-3x^2y-2xy^2)-(x^3-2xy^2+y^3)+(-x^3+3x^2y-y^3)
=2x³-3x²y-2xy²-x³+2xy²-y³-x³+3x²y-y³
=-2y³
最后的值,与x无关
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 so x抄错结果也正确
(2x^3-3x^2y-2xy^2)-(x^3-2xy^2+y^3)+(-x^3+3x^2y-y^3)
=2x³-3x²y-2xy²-x³+2xy²-y³-x³+3x²y-y³
=-2y³
最后的值,与x无关
撒打算