已知xyz满足x/(y+z)+y/(z+x)+z/(x+y)=1,求值:x^2/y+z+
已知xyz满足x/(y+z)+y/(z+x)+z/(x+y)=1,求值:x^2/(y+z)+y^2/(x+z)+z^2/(x+y)的值....
已知xyz满足x/(y+z)+y/(z+x)+z/(x+y)=1,求值:x^2/(y+z)+y^2/(x+z)+z^2/(x+y)的值.
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因为x/(y+z) + y/(z+x) +z/(x+y)=1
所以x/(y+z)=1-y/(z+x)-z/(x+y)，两边同乘以x
得x^2/(y+z)=x-xy/(z+x)-xz/(x+y)
同理y^2/(x+z)=y-xy/(z+y)-yz/(x+y)
,z^2=z-xz/(y+z)-yz/(x+z)
原式=x+y+z-(xy+zy)/(x+z)-(xz+yz)/(x+y)-(yx+zx)/(y+z)
=x+y+z-y-z-x
=0
所以x/(y+z)=1-y/(z+x)-z/(x+y)，两边同乘以x
得x^2/(y+z)=x-xy/(z+x)-xz/(x+y)
同理y^2/(x+z)=y-xy/(z+y)-yz/(x+y)
,z^2=z-xz/(y+z)-yz/(x+z)
原式=x+y+z-(xy+zy)/(x+z)-(xz+yz)/(x+y)-(yx+zx)/(y+z)
=x+y+z-y-z-x
=0
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太难了...无解...
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