数学题(因式分解):一、(x+y+z)^2-(x-y-z)^2 二、(x-1)^2(3x-2)+(2-3x)要过程

来源:学生作业帮助网 编辑:作业帮 时间:2024/11/15 10:01:09

数学题(因式分解):一、(x+y+z)^2-(x-y-z)^2 二、(x-1)^2(3x-2)+(2-3x)要过程
数学题(因式分解):一、(x+y+z)^2-(x-y-z)^2 二、(x-1)^2(3x-2)+(2-3x)
要过程

数学题(因式分解):一、(x+y+z)^2-(x-y-z)^2 二、(x-1)^2(3x-2)+(2-3x)要过程
一、(x+y+z)^2-(x-y-z)^2
=(x+y+z+x-y-z)(x+y+z-x+y+z)
=4x(y+z)
二、(x-1)^2(3x-2)+(2-3x)
=(3x-2)[(x-1)^2-1]
=x(3x-2)(x-2)

(x+y+z)^2-(x-y-z)^2 = 4x(y+z)
(x-1)^2(3x-2)+(2-3x)=3x^3-8x^2+4x

一、(x+y+z)^2-(x-y-z)^2
解,得:
==[((x+y+z)-(x-y-z)][(x+y+z)+(x-y-z)]
==(x+y+z-x+y+z)(x+y+z+x-y-z)
==4x(y+z)
二、(x-1)^2(3x-2)+(2-3x)
解,得:
==(x-1)^2(3x-2)-(3x-2)

全部展开

一、(x+y+z)^2-(x-y-z)^2
解,得:
==[((x+y+z)-(x-y-z)][(x+y+z)+(x-y-z)]
==(x+y+z-x+y+z)(x+y+z+x-y-z)
==4x(y+z)
二、(x-1)^2(3x-2)+(2-3x)
解,得:
==(x-1)^2(3x-2)-(3x-2)
==(3x-2)[(x-1)^2-1]
==(3x-2)[(x-1)^2-1]
==(3x-2)[(x-1)-1][(x-1)+1]
==(3x-2)(x-2)x

收起

一、原式=(x+y+z+(x-y-z))*(x+y+z-(x-y-z))= 4x(y+z)
二、原式=(x-1)²(3x-2)-(3x-2)=x(x-2)(3x-2)

(x+y+z)²-(x-y-z)²
=[(x+y+z)+(x-y-z)][(x+y+z)-(x-y-z)]
=2x(2y+2z)
=4x(y+z)

(x-1)²(3x-2)+(2-3x)
=(x-1)²(3x-2)-(3x-2)
=(3x-2)[(x-1)²-1]
=(3x-2)(x-1+1)(x-1-1)
=x(3x-2)(x-2)