10. Given $f(x)=\frac{1+x}{2-x}$, for $n \in \mathbf{N}_{+}$, define: $f_{1}(x)=f(x), f_{n+1}(x)=$ $f\left[f_{n}(x)\right]$. If $f_{13}(x)=f_{31}(x)$, then the analytical expression of $f_{16}(x)$ is $\qquad$ .
10. $f_{16}(x)=\frac{x-1}{x}$.

Since $f^{-1}(f(x))=x$, we have $f^{-1}\left(f_{n+1}(x)\right)=f_{n}(x)$. Given $f_{13}(x)=f_{31}(x)$, then $f_{12}(x)=f_{30}(x)$, similarly $f_{11}(x)=f_{29}(x), \cdots, f_{1}(x)=f_{19}(x), x=f_{18}(x), f^{-1}(x)=f_{17}(x)$, $f^{-1}\left(f^{-1}(x)\right)=f_{16}(x)$. Also, $f^{-1}(x)=\frac{2 x-1}{x+1}$, so $f_{16}(x)=f^{-1}\left(f^{-1}(x)\right)=\frac{x-1}{x}$.