(x-y)[(x+y)^2-xy]+(x-y)[(x-y)^2+xy]
来源:学生作业帮 编辑:百度作业网作业帮 分类:数学作业 时间:2024/07/18 05:46:35
(x-y)[(x+y)^2-xy]+(x-y)[(x-y)^2+xy]
![(x-y)[(x+y)^2-xy]+(x-y)[(x-y)^2+xy]](/uploads/image/z/15729488-8-8.jpg?t=%28x-y%29%5B%28x%2By%29%5E2-xy%5D%2B%28x-y%29%5B%28x-y%29%5E2%2Bxy%5D)
解
原式
=(x-y)[(x+y)²-xy+(x-y)²+xy]
=(x-y)[(x+y)²+(x-y)²]
=(x-y)(x²+2xy+y²+x²-2xy+y²)
=(x-y)(2x²+2y²)
=2(x-y)(x²+y²)
再问: 为什么把y=2007抄成y=2070不影响结果?
再答: (x-y)[(x+y)^2-xy]+(x-y)[(x-y)^2+xy] 这个会不会抄错,是不是(x-y)[(x+y)²-xy]+(x+y)[(x-y)²+xy] 如果没抄错,确实与y值有关啊
原式
=(x-y)[(x+y)²-xy+(x-y)²+xy]
=(x-y)[(x+y)²+(x-y)²]
=(x-y)(x²+2xy+y²+x²-2xy+y²)
=(x-y)(2x²+2y²)
=2(x-y)(x²+y²)
再问: 为什么把y=2007抄成y=2070不影响结果?
再答: (x-y)[(x+y)^2-xy]+(x-y)[(x-y)^2+xy] 这个会不会抄错,是不是(x-y)[(x+y)²-xy]+(x+y)[(x-y)²+xy] 如果没抄错,确实与y值有关啊