x+2y+z=1 x+y+2z=2 2x+y+z=5
x/2=y/3=z/5 x+3y-z/x-3y+z
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=1;x+3y+7z=-1;z+5y+8z=-2
(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+2y+z=1 x+y+2z=2 2x+y+z=5
x+y+z=2 x-3y+2z=1 2x+2y+z+5 要具体步骤..
解方程组2x+y-z=7 x+y+z =1 2x-y-z=5
3x+2y+z=14 ,x+y+z=10,z+2x+3y=1 5