{2x+z=1,3x-y=1,x+2y-3z=8

{x+y+z=1;x+3y+7z=-1;z+5y+8z=-2 试证明(x+y-2z)+(y+z-2x)+(z+x-2y)=3(x+y-2z)(y+z-2x)(z+x-2y) x+2y=3 x+y+z=36 2x+y+z=15 2y=3z x-y=1 x+2y+z x-z=-1 2x+z-y=1 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= 如果,根号x-3+| y-2 |+z^2=2z-1 求 （x+z)^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 x+2y-z=1 3x-3y+z=2 3x+3y+z=7 2x-2y+z=0 2x+y-z=1 x+3y-2z=1 解方程组：x-2y+z=-1,x+y+z=2,x+2y+3z=-1 3x+2y+z=14 x+y+2z+-3 2x+3y-z=1 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]