{2x-2y+z=0,2x+y-z+1,x+3y-2z=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 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 1.已知x,y,z满足2│x-y│+(根号2y-z)+z平方-z+(1/4)=0,求x,y,z值. {x+y+z=4 {2y+z=0 {3x+2y-4z=-1 若x-2y+4z=1,x+3y-7z=0,则y:z=?