x^2-3y^2-8z^2+2xy+2xz+14yz的因式分解
来源:学生作业帮助网 编辑:作业帮 时间:2024/11/28 07:15:42
x)35UqFF@\mhRY|Vbb"}TِjupkըNN2+S4vӉɶF`Vv
lkh$CZ@l\鉶
IF
@
):^A&`Wݣ_\g; Et
x^2-3y^2-8z^2+2xy+2xz+14yz的因式分解
x^2-3y^2-8z^2+2xy+2xz+14yz的因式分解
x^2-3y^2-8z^2+2xy+2xz+14yz的因式分解
设x^2-3y^2-8z^2+2xy+2xz+14yz=(x+ay+bz)(x+cy+dz)
则
a+c=2
b+d=2
ad+bc=14
ac=-3
bd=-8
解得a=3 b=-2 c=-1 d=4
所以x^2-3y^2-8z^2+2xy+2xz+14yz=(x-y+4z)(x+3y-2z)
24xy^2z^2(x+y+z)-32xyz(z-x-y)^2+8xyz^3(z-x-y)
24xy^2z^2(x+y+z)-32xyz(z-x-y)^2+8xyz^3(z-x-y)
代数 x^-y^/ x^-(y-z)^ 除以 x^+2xy+y^/(x-y)^-z^ 乘 x^+xy-xz/x^-xy
化简(y-x)(z-x)/(x-2y+z)(x+y-2z)+(z-y)(x-y)/(xy-2z)(y+z-2x)+(x-z)(y-z)/(y+z-2x)(x-2y+z)速速回答
化简x^2-yz/[x^2-(y+z)x+yz]+y^2-zx/[y^2-(z+x)y+zx]+z^2-xy/[z^2-(x+y)z+xy]
(2X+Z-Y)/(X^2-XY+XZ-YZ)-(Y-Z)/(X^2-XY-XZ+YZ)
若x+y=6,xy=-8,求代数式(x+y-z)^2+(x+y-z)(x+y+z)-2·z(x-y)
若x-y=6,xy=-8,求代数式(x+y+z)²+(x-y-z)(x-y+z)-2·z(x+y)的值
求证:(x^3/x+y)+(y^3/y+z)+(z^3/z+x)大于等于(xy+yz+zx)/2
(xy-1)^2+(x+y-z)(x+y-zxy)因式分解
x²-y²+5x+3y+4(x+y-2xy)(x+y-z)+(xy-1)²
24xy^2*z^2(x+y-z)-32xyz(z-x-y)^2+8xyz^3(z-x-y)
若|x-3|+|y+2|+|2z+1|=0,求(xy-yz)(y-x+z)
化简(2x-y-z/x^2-xy-xz+yz)+(2y-x-z/y^2-xy-yz+xz)+(2x-x-y/z^2-xz-yz+xy)
(x^2+yz)/(x^2+x(y-z)-yz)+(y^2-xz)/(y^2+y(x+z)+zx)+(z^2+yx)/(x^2-z(x-y)-yx)最后一项应为:(z^2-xy)/(z^2+z(y-x)-xy)
已知方程组x+2y-z=0,2x-y+8z=0,求x*x+y*y+z*z/xy+yz的值
已知x>0,y>0,z>0,证明x^3/(x+y)+y^3/(y+z)+z^3/(z+x)≥(xy+xz+yz)/2
已知x,y,z∈(0,+∞)求证:√x^+xy+y^+√y^+yz+z^+√z^+zx+x^>=3/2(x+y+z )