有这样一道题:计算(2x^4-4x^3y-2x^2y^2)-(x^4-2x^2y^2+y^3)+(-x^4+4x^3y-y^3)的值,其中x=2010,y=-1.甲同

2024-12-29 02:05:22
推荐回答(2个)
回答1:

(2x^4-4x^3y-2x^2y^2)-(x^4-2x^2y^2+y^3)+(-x^4+4x^3y-y^3)

=2x^4-4x^3y-2x^2y^2-x^4+2x^2y^2-y^3-x^4+4x^3y-y^3

=2x^4-x^4-x^4+4x^3y-4x^3y-2x^2y^2+2x^2y^2-y^3-y^3

=4x^3y-4x^3y-2x^2y^2+2x^2y^2-y^3-y^3

=-2x^2y^2+2x^2y^2-y^3-y^3

=-y^3-y^3

=-2y^3

=-2*(-1)^3

=-2*(-1)

=2

化简结果与x无关,

回答2:

原式=2x^4-4x^3y-2x^2y^2-x^4-2x^2y^2-y^3-x^4+4x^3y-y^3=2×y^3
所以结果与x没关系