Problem 1. Solve the equation $[x]+\frac{1}{[x]}=\{x\}+\frac{1}{\{x\}}$ (where $[x]$ is the integer part of $x$, and $\{x\}$ is the fractional part of $x$ ).
Solution and grading: The conditions of existence are $[x] \neq 0$ and $\{x\} \neq 0$, and we write the equation in the form $([x]-\{x\})\left(1-\frac{1}{[x] \cdot\{x\}}\right)=0$

![](https://cdn.mathpix.com/cropped/2024_06_07_e36118bc52131d497f6eg-2.jpg?height=120&width=1221&top_left_y=846&top_left_x=585)

![](https://cdn.mathpix.com/cropped/2024_06_07_e36118bc52131d497f6eg-2.jpg?height=66&width=1670&top_left_y=996&top_left_x=154)
II. $[x] \cdot\{x\}=1$. Since $\{x\} \in [0,1) \Rightarrow [x] > 1 \Rightarrow [x] = n, \{x\} = \frac{1}{n}, n \in \mathbb{N}, n \geq 2$ $-2 \text{p}$

Since $x = [x] + \{x\}$, we obtain that the set of solutions is $S = \left\{ n + \frac{1}{n} \mid n \in \mathbb{N}, n \geq 2 \right\}$ $-2 \text{p}$