2.183. $\frac{x \cdot|x-3|+x^{2}-9}{2 x^{3}-3 x^{2}-9 x}$.
Solution.

Domain of definition: $2 x^{3}-3 x^{2}-9 x \neq 0, x\left(2 x^{2}-3 x-9\right) \neq 0,2 x\left(x+\frac{3}{2}\right)(x-3) \neq 0$,

or $\left\{\begin{array}{l}x \neq 0, \\ x \neq-\frac{3}{2}, \\ x \neq 3 .\end{array}\right.$

$$
\begin{aligned}
& \frac{x \cdot|x-3|+x^{2}-9}{2 x^{3}-3 x^{2}-9 x}=\frac{x \cdot|x-3|+x^{2}-9}{x\left(2 x^{2}-3 x-9\right)}= \\
& =\left\{\begin{array}{l}
\frac{-x(x-3)+x^{2}-9}{x\left(2 x^{2}-3 x-9\right)}=\frac{-x^{2}+3 x+x^{2}-9}{2 x\left(x+\frac{3}{2}\right)(x-3)}=\frac{3(x-3)}{x(2 x+3)(x-3)}=\frac{3}{x(2 x+3)} \\
\text { for } x \in\left(-\infty ;-\frac{3}{2}\right) \cup\left(-\frac{3}{2} ; 0\right) \cup(0 ; 3) \\
\frac{x(x-3)+x^{2}-9}{x\left(2 x^{2}-3 x-9\right)}=\frac{2 x^{2}-3 x-9}{x\left(2 x^{2}-3 x-9\right)}=\frac{1}{x} \text { for } x \in(3 ; \infty) .
\end{array}\right.
\end{aligned}
$$

Answer: $\frac{3}{x(2 x+3)}$, if $x \in\left(-\infty ;-\frac{3}{2}\right) \cup\left(-\frac{3}{2} ; 0\right) \cup(0 ; 3) ; \frac{1}{x} ;$ if $x \in(3 ; \infty)$.