\(\dfrac{{{{\left( {2x + 3} \right)}^2}{{\left( {x-3} \right)}^4}}}{{\left( {1-x} \right){{\left( {x + 5} \right)}^3}}} \le 0.\)
Решим неравенство методом интервалов. Найдём нули числителя:
\({\left( {2x + 3} \right)^2}{\left( {x-3} \right)^4} = 0\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{{{\left( {2x + 3} \right)}^2} = 0,\,\,\,}\\{{{\left( {x-3} \right)}^4} = 0\,\,\,\,\,\,\,}\end{array}} \right.\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{x = -\frac{3}{2},\,\,\,}\\{x = 3.\,\,\,\,\,\,\,}\end{array}} \right.\)
Найдём нули знаменателя:
\(\left( {1-x} \right){\left( {x + 5} \right)^3} = 0\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{1-x = 0,\,\,\,\,\,}\\{{{\left( {x + 5} \right)}^3} = 0}\end{array}} \right.\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{x = 1,\,\,\,\,\,}\\{x = -5.}\end{array}} \right.\)
Следовательно, решение исходного неравенства: \(x\, \in \,\left( {-\infty ;-5} \right) \cup \left\{ {-\dfrac{3}{2}} \right\} \cup \left( {1; + \infty } \right).\)
Ответ: \(\left( {-\infty ;\;-5} \right) \cup \left\{ {-\dfrac{3}{2}} \right\} \cup \left( {1;\; + \infty } \right).\)