x+y+z=2,x-z=-2,y+z=1
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
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)^2-(x-y+z)^2=?
(x+y-z)(x-y+z)=
{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
分解因式:f(x,y,z)=x^2(y-z)+y^2(z-x)+z^2(x-y)
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-
x/2=y/3=z/5 x+3y-z/x-3y+z
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.