x=2 2y=z xyz=1+9 y=? z=?

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=2 2y=z xyz=1+9 y=? z=? 先化简再求值3xyz+2(x^2y+y^2z-xyz)-xyz+2z^2x x=1 y= -1 z=2 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= 已知x^2+y^2+z^2=1,求证x+y+z-2xyz x+y+z=1 求xyz/(x+y)(y+z)(z+x)的最大值 已知x,y,z都是正数,且xyz=1,求证：x^2/(y+z)+y^2/(x+z)+z^2/(x+y)≥3/2 已知x+y-z/z=x-y+z/y=-x+y+z/x,且xyz不等于0,求分式[(x+y)(x+z)(y+z)]/xyz 已知实数xyz满足x/y+z+y/z+x+z/x+y=1求x^2/y+z+y^2/z+x+z^2/x+y的值 2x+y+z=17 x+2z+y=14 x+z+2y=13 求:xyz 设X+Y+Z=0求X^3+X^2Z-XYZ+Y^2Z+Y^3的值 x*x+y*y+2z*z-2x+4y+4z+7=0,求xyz的值