2.180. $\frac{|x-1|+|x|+x}{3 x^{2}-4 x+1}$.
Solution.

Domain of definition: $\left\{\begin{array}{l}x \neq \frac{1}{3}, \\ x \neq 1 .\end{array}\right.$

$$
\frac{|x-1|+|x|+x}{3 x^{2}-4 x+1}=\left\{\begin{array}{l}
\frac{-x+1-x+x}{(x-1)(3 x-1)}=-\frac{x-1}{(x-1)(3 x-1)}=\frac{1}{1-3 x} \text { for } x \in(-\infty ; 0) \\
\frac{-x+1+x+x}{(x-1)(3 x-1)}=\frac{x+1}{(x-1)(3 x-1)} \text { for } x \in\left[0 ; \frac{1}{3}\right) \cup\left(\frac{1}{3} ; 1\right) ; \\
\frac{x-1+x+x}{(x-1)(3 x-1)}=\frac{3 x-1}{(x-1)(3 x-1)}=\frac{1}{x-1} \text { for } x \in(1 ; \infty) .
\end{array}\right.
$$

Answer: $\frac{1}{1-3 x}$ for $x \in(-\infty ; 0) ; \frac{x+1}{(x-1)(3 x-1)}$ for $x \in\left[0 ; \frac{1}{3}\right) \cup\left(\frac{1}{3} ; 1\right)$; $\frac{1}{x-1}$ for $x \in(1 ; \infty)$