神马品牌网夜尊异世男主:两道计算题：
来源：百度文库 编辑：神马品牌网 时间：2024/05/13 19:35:57
（6x^3-4x^2)÷(-2x^2)-2(x+1)
(x+y-z)(x-y+z)-(x+y+z)(x-y-z)
(x+y-z)(x-y+z)-(x+y+z)(x-y-z)
1.原式=2x^2(3x-2)÷2x^2-2x-2
=3x-2-2x-2
=x-4
2.原式=[x+(y-z)][x-(y-z)]-[x+(y+z)][x-(y+z)]
=x^2-(y-z)^2-[x^2-(y+z)^2]
=(y+z)^2-(y-z)^2
=y^2+2yz+z^2-y^2+2yz-z^2
=4yz
（6x^3-4x^2)÷(-2x^2)-2(x+1) =-5X 只需约分，化简
(x+y-z)(x-y+z)-(x+y+z)(x-y-z)=4YZ 可以把前后两项都写成平方的差，容易消去
不知