Example 2. Given $x=\frac{\sqrt{5}+1}{2}$. Then what is the value of $\frac{x^{3}+x+1}{x^{5}}$?
Given that $\frac{1}{x}=\frac{\sqrt{5}-1}{2}$, then $x-\frac{1}{x}=1$. Therefore,
$$
\begin{array}{l}
\frac{x^{3}+x+1}{x^{5}}=\frac{x^{2}+1+\frac{1}{x}}{x^{4}} \\
=\frac{x^{2}+\left(x-\frac{1}{x}\right)+\frac{1}{x}}{x^{4}}=\frac{x^{2}+x}{x^{4}} \\
=\frac{1+\frac{1}{x}}{x^{2}}=\frac{\left(x-\frac{1}{x}\right)+\frac{1}{x}}{x^{2}}=\frac{1}{x} \\
=\frac{\sqrt{5}-1}{2} .
\end{array}
$$

Note: Here, we first constructed the expression $x-\frac{1}{x}$ which equals 1, and then substituted $x-\frac{1}{x}$ for the constant 1 in the required expression.