2x+y+z=4 x+2y+z=8 x +y+2z=24
试证明(x+y-2z)+(y+z-2x)+(z+x-2y)=3(x+y-2z)(y+z-2x)(z+x-2y)
2x+y+z=4 x+2y+z=8 x +y+2z=24
(x+y-z)^2-(x-y+z)^2=?
x=y/z=z/3,x+y+z =12,求2x+3y+4z是多少,
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/2=y/3=z/5 x+3y-z/x-3y+z
(4x-2y-z)-{5x[8y-2y-(x+y)]-x+(3y-10z)]=? kuai
已知x+4y+z=24,2x+7y=2z=41,求x+y+z
x+y+z=51 4x+8y+5z=300 x+y+2z=67
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