\(\left( {x-1} \right)\left( {x + 3} \right)\left( {2x-7} \right)\left( {3-x} \right) \le 0.\)
Решим неравенство методом интервалов:
\(\left( {x-1} \right)\left( {x + 3} \right)\left( {2x-7} \right)\left( {3-x} \right) = 0\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{x-1 = 0,\,\,\,\,}\\{x + 3 = 0,\,\,\,}\\{2x-7 = 0,}\\{3-x = 0\,\,\,\,}\end{array}} \right.\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{x = 1,\,\,\,}\\{x = -3,}\\{x = \dfrac{7}{2},\,}\\{x = 3.\,\,\,}\end{array}} \right.\)
Следовательно, решение исходного неравенства: \(x\, \in \,\left( {-\infty ;-3} \right] \cup \left[ {1;3} \right] \cup \left[ {\dfrac{7}{2}; + \infty } \right).\)
Ответ: \(\left( {-\infty ;\;-3} \right] \cup \left[ {1;\;3} \right] \cup \left[ {\dfrac{7}{2};\; + \infty } \right).\)