2.185. $\frac{x^{2}-1+|x+1|}{|x| \cdot(x-2)}$.

2.185. $\frac{x^{2}-1+|x+1|}{|x| \cdot(x-2)}$.
## Solution.

odZ: $\left\{\begin{array}{l}x \neq 0, \\ x \neq 2 .\end{array}\right.$

$$
\frac{x^{2}-1+|x+1|}{|x| \cdot(x-2)}=
$$

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

Answer: $-\frac{x+1}{x}$, if $x \in(-\infty ;-1) ; \frac{x+1}{2-x}$, if $x \in[-1 ; 0) ; \frac{x+1}{x-2}$, if $x \in(0 ; 2) \cup(2 ; \infty)$.