中科新闻网化妆品封闭剂:数学题（初二）
来源：百度文库 编辑：中科新闻网 时间：2024/04/29 15:57:33
1.分解因式：
（x-y-z)(x+y-z)-(y-z-x)(y+z-x)
哪个答案才对啊？
（x-y-z)(x+y-z)-(y-z-x)(y+z-x)
哪个答案才对啊？
(x-y-z)(x+y-z)-(y-z-x)(y+z-x)
=(x-z-y)(x-z+y)-(y-x-z)(y-x+z)
=(x-z)^2-y^2-(y-x)^2+z^2
=(x-z)^2-(y-x)^2+z^2-y^2
=(x-z+y-x)(x-z-y+x)+(z-y)(z+y)
=(y-z)(2x-z-y)-(y-z)(z+y)
=2(y-z)(x-y-z)
这个是正确答案！
是不是这样：
(x-y-z)(x+y-z)-(y-z-x)(y+z-y)
=(x+y-z)(x-y-z-y+z+x)
=(x+y-z)(2(x-y))
=2(x-y)(x+y-z)
=(x-z-y)(x-z+y)-(y-x-z)(y-x+z)
=(x-z)^2-y^2-(y-x)^2+z^2
=(x-z)^2-(y-x)^2+z^2-y^2
=(x-z+y-x)(x-z-y+x)+(z-y)(z+y)
=(y-z)(2x-z-y)-(y-z)(z+y)
=2(y-z)(x-y-z)
=2[x(y-x)+(z-y)(z+y)]
2(xy-x*x+z*z+y*y)
设x-z为a,y-x为b
(a-y)(a+y)-(b-z)(b+z)=a^2-y^2-b^2+z^2=
2z^2-2y^2+2xy-2xz=2(z-y)(z+y-x)
答案是2(z-y)(z+y-x)
我的回答：
（x-y-z)(x+y-z)-(y-z-x)(y+z-x)
=[x-(y+z)][x+(y-z)]-[(y-z)-x][(y+z)-x]
=[x-(y+z)][x+(y-z)]-{-[-(y+z)+x]}{-[(y-z)+x]}
=[x-(y+z)][x+(y-z)]-[x-(y+z)][x+(y-z)]
=0