2道因式分解,（x+1）（x+2)(x+3)(x+4)-24 (x+y-z)(x-y+z)-4yz

来源：学生作业帮编辑：作业帮分类：数学作业时间：2024/05/10 08:29:34

2道因式分解,
（x+1）（x+2)(x+3)(x+4)-24
(x+y-z)(x-y+z)-4yz

（x+1）（x+2)(x+3)(x+4)-24
=(x+1)(x+4)(x+2)(x+3)-24
=（x^2+5x+4)(x^2+5x+6)-24
=(x^2+5x+5-1）（x^2+5x+5+1）-24
=（x^2+5x+5)^2-1-24
=(x^2+5x+5+5)(x^2+5x+5-5)
=x(x+5)(x^2+5x+10)
(x+y-z)(x-y+z)-4yz
=[x+(y-z)][x-(y-z)]-4yz
=x^2-(y-z)^2-4yz
=x^2-y^2-z^2+2yz-4yz
=x^2-(y+z)^2
=(x+y+z)(x-y-z)

因式分解 (x+y+z)^2+yz(y+z)+xyz 因式分解x²-y²-z²+2yz (x+2y-7z)^3+(3x-4y+6z)^3-(4x-2y-z)^3 因式分解已知三个数x,y,z,满足xy/x+y=-2，yz/y+z=4/3,zx/z+x=-4/3,求(xyz)/(xy+yz+ (2X+Z-Y)/(X^2-XY+XZ-YZ)-(Y-Z)/(X^2-XY-XZ+YZ) 因式分解x²-y²-z²=2yz+2x+1 因式分解x^2+y^2+z^2+2xy+2yz+2xz-2x-2y-2z+1 因式分解:(x+y+z)的平方-4(x-y-z) xyz-xy-xz+x-yz+y+z-1因式分解已知x/2=y/3=z/4,求下列各式（1）(x+y+z)/x (2)(x-y+2x/(x-y-2z) xy／x+y＝1,yz／y＋z＝2,xz／x＋z＝3求x,y （1）因式分解：x的平方-y的平方-2x-4y-3 （2）因式分解：x的平方-4y的平方-9z的平方-12yz 谢谢!