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)

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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)
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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)
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)

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)
由合分比定理知x/(y+z+u)=y/(z+u+x)=z/(u+x+y)=u/(x+y+z)=(x+y+z+u)/(3x+3y+3z+3u)=1/3
化简得四元一次方程组,仅三个独立,得x=y=z=u
代入问题可得结果.