{x-2y-z=6 2x+y-3z=1 2x-y+2z=10
试证明(x+y-2z)+(y+z-2x)+(z+x-2y)=3(x+y-2z)(y+z-2x)(z+x-2y)
x/2=y/3=z/5 x+3y-z/x-3y+z
如果|x+y+z-6|+|2x+3y-z-12|+|2x-y-z|=0求x,y,
三元一次方程组数学题x+2y+2z=33x+y-2z=72x+3y-2z=10x-y=2z-x=3y+z=-1x-y-z
已知 x,y,z都是正实数,且 x+y+z=xyz 证明 (y+x)/z+(y+z)/x+(z+x)/y≥2(1/x+1
x+y+z=4 2x+3y-z=6 3x+2y+2z=10
3x+2y+z=14 ,x+y+z=10,z+2x+3y=1 5
[3x+2y+z=14,x+y+z=10,2x+3y-z=1]
{2x+3y-4z=-5 x+y+z=6 x-y+3z=10
3道高数题,1,函数F(x,y,z)=(e^x) * y * (z^2) ,其中z=z(x,y)是由x+y+z+xyz=
{x+y+z=1;x+3y+7z=-1;z+5y+8z=-2
如果,根号x-3+| y-2 |+z^2=2z-1 求 (x+z)^y