\(\dfrac{{\left( {{x^3}-2{x^2}-3x} \right)\left( {x-3} \right)}}{{x\left( {x + 1} \right)\left( {2-x} \right)}} \ge 0\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\dfrac{{x\left( {{x^2}-2x-3} \right)\left( {x-3} \right)}}{{-x\left( {x + 1} \right)\left( {x-2} \right)}} \ge 0\,\,\,\,\,\,\, \Leftrightarrow \)
\( \Leftrightarrow \,\,\,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{\dfrac{{\left( {x + 1} \right)\left( {x-3} \right)\left( {x-3} \right)}}{{\left( {x + 1} \right)\left( {x-2} \right)}} \le 0,}\\{x \ne 0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\end{array}} \right.\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{\dfrac{{{{\left( {x-3} \right)}^2}}}{{x-2}} \le 0,}\\{x \ne 0,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{x \ne -1.\,\,\,\,\,\,\,\,\,\,\,}\end{array}} \right.\)
Следовательно, решение исходного неравенства: \(x\, \in \,\left( {-\infty ;-1} \right) \cup \left( {-1;0} \right) \cup \left( {0;2} \right) \cup \left\{ 3 \right\}.\)
Ответ: \(\left( {-\infty ;\;-1} \right) \cup \left( {-1;\;0} \right) \cup \left( {0;\;2} \right) \cup \left\{ 3 \right\}.\)