## Problem 3.

Solve in $\mathbb{R}$ the equation: $\left[\frac{1}{10-x}\right]=\frac{1}{10-[x]}$
## Solution:

$\left.\begin{array}{l}\frac{1}{10-[x]} \in \mathbb{Z} \\ {[x] \in \mathbb{Z}}\end{array}\right\} \Rightarrow 10-[x] \mid 1 \Rightarrow 10-[x] \in\{ \pm 1\}$

I. $10-[x]=1 \Rightarrow[x]=9 \Rightarrow x \in[9,10)$

$$
\left[\frac{1}{10-x}\right]=1 \Rightarrow \frac{1}{10-x} \in[1,2) \Rightarrow x \in\left[9 ; \frac{19}{2}\right)
$$

II. $10-[x]=-1 \Rightarrow[x]=11 \Rightarrow x \in[11,12)$

$$
\begin{aligned}
& {\left[\frac{1}{10-x}\right]=-1 \Rightarrow \frac{1}{10-x} \in[-1,0) \Rightarrow x \in[11 ; 12)} \\
& x \in\left[9 ; \frac{19}{2}\right) \cup[11,12)
\end{aligned}
$$