x+y+z=24 x=y+2 x+y=z
(x+y-z)(x-y+z)=
1/x+1/y+1/z=1.x,y,z,属于不同自然数 y,
试证明(x+y-2z)+(y+z-2x)+(z+x-2y)=3(x+y-2z)(y+z-2x)(z+x-2y)
X+Y+Z=?
已知x平方+ y平方+z平方减2z+4y+6z+14=0,求x+y+z的直
解方程组:2X=3Y Y=2Z X+2Y+Z=16
f(x,y,z,w)=x*(x+y)*(x+y+z)*(x+y+z+w)
(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
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-
若x/4=y/3=z/4,且x+y+z=12,求2x+3y+4z
分解因式:f(x,y,z)=x^2(y-z)+y^2(z-x)+z^2(x-y)