x/(y+z+u)=y/(z+u+x)=z/(u+y+x)=u(x+y+z)求(x+y)/(z+u)+(y+z)/(y+x)+(z+y)/(x+y)+(u+x)/(z+y)

已知x/(y+z+u)=y/(z+u+x)=z/(u+x+y)=u/(x+y+z)求(x+y)/(z+u)+(y+z)/(x+u)+(z+u)/(x+y)+(u+x)/(y+z) x/(y+z+u)=y/(z+u+x)=z/(u+y+x)=u(x+y+z)求(x+y)/(z+u)+(y+z)/(y+x)+(z+y)/(x+y)+(u+x)/(z+y) 求全微分 u=x^y^z z=f(u) u=x/y,求x*∂z/∂x +y*z∂z/∂y 求u=x^y^Z偏导数 分别求u对x,u对y,u对z u=arctan(x-y)^z偏导数u/z只求关于Z的求导, u=x(z+y) z=sin(x+z) 求二阶偏导数σ2u／σxσy u=x(z+y) z=sin(x+y) 求二阶偏导数σ2u／σxσy u=x^y^z,求u对z的偏导数 四元一次方程的解法y+z+u=1z+u+x=2u+x+y=3x+y+z=4 设x+y+z=11求函数u=2x*x+3y*y+z*z的最小值 帮帮忙拜托啦u=ln(x^y乘y^z乘z^y)求du Mathematica 这个式子哪里错了DSolve[(y + z) D[u[x,y,z],x] + (z + x) D[u[x,y,z],y] + (x + y) D[u[x,y,z],z] == 0,u[x,y,z],{x,y,z}] 四元一次方程组,天啊.、、、、、、x+y+z+u=4x+y-z-u=-2x-y+z-u=0x+y-z+u=2 求偏导数u=x^y^z u=x^(y/z),求偏导数 X+2Y=5Y+2Z=8Z+2U=11U+2X=6 求u=x+y+z,求u在z=2x^2+y^2条件下的极值