这个数学题怎么做，要过程
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解：根据条件，则x/(y+z)=1-y/(z+x)-z/(x+y)
∴ x^2/(y+z)=x-xy/(z+x)-xz/(x+y)
同理，y^2/(z+x)=y-xy/(y+z)-yz/(x+y)，z^2/(x+y)=z-xz/(y+z)-yz/(z+x)
∴ x^2/(y+z)+y^2/(z+x)+z^2/(x+y)
=x-xy/(z+x)-xz/(x+y)+y-xy/(y+z)-yz/(x+y)z-xz/(y+z)-yz/(z+x)
=x+y+z-(xy+yz)/(z+x)-(xz+yz)/(x+y)-(xy+xz)/(y+z)
=x+y+z-y-z-x
=0
答案为：B
∴ x^2/(y+z)=x-xy/(z+x)-xz/(x+y)
同理，y^2/(z+x)=y-xy/(y+z)-yz/(x+y)，z^2/(x+y)=z-xz/(y+z)-yz/(z+x)
∴ x^2/(y+z)+y^2/(z+x)+z^2/(x+y)
=x-xy/(z+x)-xz/(x+y)+y-xy/(y+z)-yz/(x+y)z-xz/(y+z)-yz/(z+x)
=x+y+z-(xy+yz)/(z+x)-(xz+yz)/(x+y)-(xy+xz)/(y+z)
=x+y+z-y-z-x
=0
答案为：B
追问
我已经做出来了，不过还是谢谢你
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