x+2y+z=0 x-y+2z=2 2x-y+3z=-1
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 大于0,x+y+z=2,求证 xz/y(y+z)+zy/x(x+y)+yx/z(z+x)大于等于2/3
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,
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
2x-2y+z=0 2x+y-z=1 x+3y-2z=1
{2x-2y+z=0,2x+y-z+1,x+3y-2z=1
3道高数题,1,函数F(x,y,z)=(e^x) * y * (z^2) ,其中z=z(x,y)是由x+y+z+xyz=
{x+y+z=1;x+3y+7z=-1;z+5y+8z=-2
如果,根号x-3+| y-2 |+z^2=2z-1 求 (x+z)^y
x+2y=3 x+y+z=36 2x+y+z=15 2y=3z x-y=1 x+2y+z x-z=-1 2x+z-y=1
已知(x+y+z)^2=x^2+y^2+z^2,证明x(y+z)+y(z+x)+z(x+y)=0