qwe {2x+y+z=2x+2y+z=4x+y+2z=6

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、z满足方程组：x+y-z=6；y+z-x=2；z+x-y=0 求x、y、z的值 (x+y-z)^2-(x-y+z)^2=? x,y,z为实数 且(y-z)^2+(x-y)^2+(z-x)^2=(y+z-2x)^2+(x+z-2y)^2+(x+y 分解因式:f(x,y,z)=x^2(y-z)+y^2(z-x)+z^2(x-y) x,y,z为实数且(y-z)平方+(x-y)平方+(z-x)平方=(y+z-2x)平方+(z+x-2y)平方+(x+y- x=y/z=z/3,x+y+z ＝12,求2x+3y+4z是多少, 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, 1.x+y=16,y+z=12,z+x=102.3x-y+z=4,2x+3y-z=12,x+y+z=63.x+y+z=6 2x+y+z=2 x+2y+z=4 x+y+2z=6 x+y+z=4 2x+3y-z=6 3x+2y+2z=10