Solve the following equation.

$$
\frac{1}{[x]}+\frac{1}{\{x\}}=x
$$

( $[x]$ denotes the integer part of the number $x$, and $\{x\}$ denotes the fractional part of $x$.)
As $x \neq 0$, otherwise there would be a 0 in the denominators. Bring the left side of the equation to a common denominator:

$$
\frac{\{x\}+[x]}{\{x\} \cdot[x]}=x
$$

In the numerator, $x$ stands (the fractional part and the integer part of the number give the number itself). Simplifying with $x \neq 0$ and rearranging, we get $\{x\} \cdot[x]=1$. Since $\{x\}$ is never negative, $[x]>0$.

Let $[x]=n>1$ be an integer, then according to the equation $\{x\}=\frac{1}{n}$, so $x=[x]+\{x\}=n+\frac{1}{n}$. It is easy to see that if $n$ is any integer greater than 1, then $0 < \{x\} < 1$. Therefore, the solutions are all numbers of the form $n + \frac{1}{n}$, where $n$ is an integer greater than 1.