\(\left( {2x + 1} \right)\left( {4-x} \right)\left( {x-2} \right) > 0.\)
Решим неравенство методом интервалов:
\(\left( {2x + 1} \right)\left( {4-x} \right)\left( {x-2} \right) > 0\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{2x + 1 = 0,}\\{4-x = 0,\,\,}\\{x-2 = 0\,\,\,}\end{array}} \right.\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{x = -\dfrac{1}{2},}\\{x = 4,\,\,\,\,}\\{x = 2.\,\,\,\,}\end{array}} \right.\)
Следовательно, решение исходного неравенства: \(\left( {-\infty ;-\dfrac{1}{2}} \right) \cup \left( {2;4} \right).\)
Ответ: \(\left( {-\infty ;\;-\dfrac{1}{2}} \right) \cup \left( {2;\;4} \right).\)