Problem 1. Solve in $\boldsymbol{R}$ the equation:

$$
\left[\frac{4 x+1}{6}\right]=\left\{\frac{2 x-1}{3}\right\}+x
$$
Solution:

$\left\{\frac{2 x-1}{3}\right\}=\frac{2 x-1}{3}-\left[\frac{2 x-1}{3}\right]$

$\mathbf{1 p}$

The equation becomes $\left[\frac{2 x-1}{3}\right]+\left[\frac{4 x+1}{6}\right]=x+\frac{2 x-1}{3}$

According to Hermite's identity $\left[\frac{2 x-1}{3}\right]+\left[\frac{4 x+1}{6}\right]=\left[2 \cdot \frac{2 x-1}{3}\right] \Rightarrow\left[\frac{4 x-2}{3}\right]=\frac{5 x-1}{3}$

Let $k \in Z, k=\left[\frac{4 x-2}{3}\right] \Rightarrow x=\frac{3 k+1}{5}$

## 1p

$\left.\begin{array}{l}k \leq \frac{(4 x-2)}{3}<k+1 \Leftrightarrow k \in(-7,-2] \\ k \in Z\end{array}\right\} \Rightarrow k \in\{-6 ;-5 ;-4 ;-3 ;-2\}$

$\Rightarrow x \in\left\{\frac{-17}{5}, \frac{-14}{5}, \frac{-11}{5}, \frac{-8}{5},-1\right\}$

1 .