\(x\left(x-10\right)+x-10=0\)

\(\Leftrightarrow x\left(x-10\right)+1\left(x-10\right)=0\)

\(\Leftrightarrow\left(x-10\right)\left(x+1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-10=0\\x+1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=10\\x=-1\end{cases}}\)

\(\frac{x}{y+z}+\frac{y}{x+z}+\frac{z}{x+y}=1\Leftrightarrow\left(x+y+z\right)\left(\frac{x}{y+z}+\frac{y}{x+z}+\frac{z}{x+y}\right)=x+y+z\)

<=>\(\frac{x^2+x\left(y+z\right)}{y+z}+\frac{y^2+y\left(z+x\right)}{z+x}+\frac{z^2+z\left(x+y\right)}{x+y}=x+y+z\)

<=>\(\frac{x^2}{y+z}+x+\frac{y^2}{z+x}+y+\frac{z^2}{x+y}+z=x+y+z\)

<=>\(S=\frac{x^2}{y+z}+\frac{y^2}{x+z}+\frac{z^2}{x+y}=0\)

 x/(y+z)+y/(x+z)+z/(x+y)=1

=>\(\frac{x^2}{\left(y+z\right)^2}\)+\(\frac{y^2}{\left(x+z\right)^2}\)+\(\frac{z^2}{\left(x+y\right)^2}\)+2(\(\frac{xy}{\left(y+z\right)\cdot\left(x+z\right)}\)+\(\frac{yz}{\left(x+z\right)\left(x+y\right)}\)+\(\frac{zx}{\left(z+y\right)\cdot\left(x+y\right)}\))=1

\(x^3-3x+1=\frac{1}{2}x^2\left(2x-1\right)+\frac{1}{4}x\left(2x-1\right)-\frac{11}{8}\left(2x-1\right)-\frac{3}{8}\)

\(=\left(2x-1\right)\left(\frac{1}{2}x^2+\frac{1}{4}x-\frac{11}{8}\right)-\frac{3}{8}\)