## SUBJECT 1.

Solve in the set of real numbers the equations:
a) $x-2 \sqrt{x-2}+\left[x-\frac{2010}{671}\right]^{2}=1$
b) $\left[\left\{\frac{x-2}{3}\right\}-\frac{x-2}{3}\right]=\frac{1-3 x}{5}$,

where $[\mathrm{a}]$ represents the integer part of the number a and $\{\mathrm{b}\}$ represents the fractional part of the number $b$.
Subject 1. a) The equation is equivalent to $(\sqrt{x-2}-1)^{2}+\left[x-\frac{2010}{671}\right]^{2}=0$

Thus $\sqrt{x-2}=1$ and $\left[x-\frac{2010}{671}\right]=0$

The solution is $x=3$

b) $\left[\left\{\frac{x-2}{3}\right\}-\frac{x-2}{3}\right]=\frac{1-3 x}{5} \Leftrightarrow\left[-\left[\frac{x-2}{3}\right]\right]=\frac{1-3 x}{5} \Leftrightarrow-\left[\frac{x-2}{3}\right]=\frac{1-3 x}{5} \Leftrightarrow\left[\frac{x-2}{3}\right]=\frac{3 x-1}{5} .$.

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

Therefore $k \leq \frac{x-2}{3}<k+1 \Leftrightarrow 3 k \leq \frac{5 k+1}{3}-2<3 k+3 \Leftrightarrow k \in\{-2,-1\}$

Thus the solution is $x \in\left\{-3,-\frac{4}{3}\right\}$