Solve the equation
$$
\frac{x}{2+\frac{x}{2+} \cdot \cdot 2+\frac{x}{1+\sqrt{1+x}}}=1 .
$$
(The expression on the left side of the equation contains 1985 twos)
(1985 All-Soviet Union Mathematical Olympiad)
Solve: Since $\frac{x}{\sqrt{x+1}+1}=\sqrt{x+1}-1$, let $f(x)=2+\sqrt{x+1}-1$ $=\sqrt{x+1}+1$, then the original equation is clearly an iterative equation
$$
\begin{array}{c}
f^{[n]}(x)=2+\frac{f^{[n-1]}(x)}{\sqrt{f^{[n-1]}(x)+1}+1}, \\
f^{[n]}(x)=\sqrt{f^{[n-1]}(x)+1}+1 .
\end{array}
$$
The fixed points of $f(x)$ are the roots of $\sqrt{x+1}+1=x$, solving which gives $x_{1}=0, x_{2}=3$. $x_{0}$ is obviously not a solution, hence $x=3$.