5. Solve the equation:

$$
x+\frac{7}{x}=[x]+\frac{7}{[x]}
$$

where $x=[x]+\{x\}$.
Solution. After equivalent transformations, we arrive at the equation

$$
(x-[x])\left(1-\frac{7}{x[x]}\right)=0
$$

which is equivalent to the system of equations $x-[x]=0$ and $[x]=\frac{7}{x}$. The solutions to the first equation are all non-zero integers. In the second equation, express the integer part of the number as $[x]=x-\{x\}$, from which $\{x\}=x-\frac{7}{x}$, which implies that $0<\mathrm{x}-\frac{7}{\mathrm{x}}<1$ when $[\mathrm{x}] \neq 0$. The solutions to the double inequality are $\mathrm{x} \in\left[-\sqrt{7} ; \frac{1-\sqrt{29}}{2}\right) \cup\left[\sqrt{7} ; \frac{1+\sqrt{29}}{2}\right)$ and, consequently,
$[x] \in\{-3 ; 2 ; 3\}$. Now, from the equation $[x]=\frac{7}{x}$, we find $x=-\frac{7}{3} ; x=\frac{7}{2} ; x=\frac{7}{3}$. From the obtained values of $x$, we select those that belong to the intervals $\left[-\sqrt{7} ; \frac{1-\sqrt{29}}{2}\right) \cup\left[\sqrt{7} ; \frac{1+\sqrt{29}}{2}\right) \cdot$ Thus, $\mathrm{x}=-\frac{7}{3}$.

Answer: $\mathrm{x} \in \mathrm{Z}$ and $\mathrm{x}=-\frac{7}{3}$.

Criteria: Full solution - 7 points; two extra roots found - 5 points; the set of integers indicated - 3 points.