一初三代数题

将等式x/(y+z)+y/(x+z)+z/(x+y)=1
两边分别乘以x,y,z,得:
x^2/(y+z)+xy/(x+z)+xz/(x+y)=x;
xy/(y+z)+y^2/(x+z)+yz/(x+y)=y;
xz/(y+z)+yz/(x+z)+z^2/(x+y)=z;
将这三个等式左右相加:
x^2/(y+z)+y^2/(x+z)+z^2/(x+y)+(xy+yz)/(x+z)+(xy+xz)/
(y+z)+(xz+yz)/(x+y)=x^2/(y+z)+y^2/(x+z)+z^2/(x+y)+x+y+z=x+y+z;
所以:x^2/(y+z)+y^2/(x+z)+z^2/(x+y)=0;

不是2么？

已知：x/(y+z)+y/(x+z)+z/(x+y)=1
求：x2/(y+z)+y2/(x+z)+z2/(x+y)的值。

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

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

化简已知得：x^3+y^3+z^3+xyz=0