x/2=y/3=z/4,2x+y-z=6 解此方程组

x+y/2=z+x/3=y+z/4 x+y+z=27 x+2y=3x+2z=4y+z 求x:y:z 3x-y+z=4,2z+3y-z=12,x+y+z=6的解 试证明(x+y-2z)+(y+z-2x)+(z+x-2y)=3(x+y-2z)(y+z-2x)(z+x-2y) 已知(x+y)(x+z)=x,(y+z)(y+x)=2y,(z+x)(z+y)=3z,求x,y,z x+y+z=6 x+2y+3z=14 y+1=z 2x+z=10 x+y-z=4 3x-y-z=0 三元一次方程x+y+z=0 4x+2y+z=3 9x+3y+z=6x+y+z=04x+2y+z=3 9x+3y+z=6 x=y/z=z/3,x+y+z ＝12,求2x+3y+4z是多少, 2x-y+2z=6 3x+y-3z=-4 x+y+z=6 x+y+z=4 2x+3y-z=6 3x+2y+2z=10 x+y-z=6 x-3y+2z=1 3x+2y-z=4 阿 x+y-z=6,x-3y+2z=1,3x+2y-z=4 X+y-z=6;x-3y+2z=0;3x+2y-z=4 x+y-z=6 x-3y+2z=1 3x+2y-z=4 2x-y+2z=-17 3x+y-3z=-4 x+y+z=6 2x-y+2z=-17 3x+y-3z=-4 x+y+z=6 x-y+z=-1 5x+2y-z=6 4x-3y+2z=-5