急急急！已知x,y,z满足x/(y+z)+y/(z+x)+z/(x+y)=1,求代数式x^2/(y+z)+y^2/(z+x)+z^2/(x+y)的值

x^2/(y+z)+y^2/(z+x)+z^2/(x+y)=2*x/(y+z)+y/(z+x)+z/(x+y)=2*1=2

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x平方/(y+z)+y平方/(z+x)+z平方/(x+y)=2[x/(y+z)+y/(z+x)+z/(x+y)]=2乘以1=2

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
所以原式=x+y+z-(xy+zy)/x+z-(xz+yz)/x+y-(yx+zx)/y+z
=x+y+z-y-z-x
=0