第2小題
易知\(x,y,z\ne 0\)
\(\begin{align}
& \frac{{{x}^{2}}}{y+z}=-\frac{{{y}^{2}}}{z+x}-\frac{{{z}^{2}}}{x+y} \\
& \frac{x}{y+z}=-\frac{{{y}^{2}}}{zx+{{x}^{2}}}-\frac{{{z}^{2}}}{{{x}^{2}}+xy} \\
& \frac{y}{z+x}=-\frac{{{x}^{2}}}{{{y}^{2}}+yz}-\frac{{{z}^{2}}}{xy+{{y}^{2}}} \\
& \frac{z}{x+y}=-\frac{{{x}^{2}}}{yz+{{z}^{2}}}-\frac{{{y}^{2}}}{{{z}^{2}}+zx} \\
& \\
& \frac{1}{{{y}^{2}}+yz}+\frac{1}{yz+{{z}^{2}}}=\frac{1}{y\left( y+z \right)}+\frac{1}{z\left( y+z \right)}=\frac{z+y}{yz\left( y+z \right)}=\frac{1}{yz} \\
& \frac{1}{{{z}^{2}}+zx}+\frac{1}{zx+{{x}^{2}}}=\frac{1}{zx} \\
& \frac{1}{{{x}^{2}}+xy}+\frac{1}{xy+{{y}^{2}}}=\frac{1}{xy} \\
& \\
& \frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{x+y}=-\left( \frac{{{x}^{2}}}{yz}+\frac{{{y}^{2}}}{zx}+\frac{{{z}^{2}}}{xy} \right) \\
& \frac{{{x}^{2}}}{yz}+\frac{{{y}^{2}}}{zx}+\frac{{{z}^{2}}}{xy}=-1 \\
\end{align}\)
