\(\dfrac{{-{x^3} + 4{x^2} + 5x}}{{x\left( {2x + 1} \right)\left( {x + 1} \right)}} \ge 0\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\dfrac{{-x\left( {{x^2}-4x-5} \right)}}{{x\left( {2x + 1} \right)\left( {x + 1} \right)}} \ge 0\,\,\,\,\,\,\, \Leftrightarrow \)
\( \Leftrightarrow \,\,\,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{\dfrac{{\left( {x + 1} \right)\left( {x-5} \right)}}{{\left( {2x + 1} \right)\left( {x + 1} \right)}} \le 0,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{x \ne 0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\end{array}} \right.\,\, \Leftrightarrow \,\,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{\dfrac{{x-5}}{{2x + 1}} \le 0,\,\,\,}\\{x \ne 0,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{x \ne -1.\,\,\,\,\,\,\,\,\,\,\,}\end{array}} \right.\)
Следовательно, решение исходного неравенства: \(x\, \in \,\left( {-\dfrac{1}{2};0} \right) \cup \left( {0;5} \right].\)
Ответ: \(\left( {-\dfrac{1}{2};\;0} \right) \cup \left( {0;\;5} \right].\)