\({\left( {x-4} \right)^2}{\left( {x + 2} \right)^2}\left( {x-7} \right) \le 0.\)
Решим неравенство методом интервалов:
\({\left( {x-4} \right)^2}{\left( {x + 2} \right)^2}\left( {x-7} \right) = 0\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{{{\left( {x-4} \right)}^2} = 0,}\\{{{\left( {x + 2} \right)}^2} = 0,}\\{x-7 = 0\,\,\,\,\,\,\,\,}\end{array}} \right.\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{x = 4,\,\,\,\,}\\{x = -2,}\\{x = 7.\,\,\,}\end{array}} \right.\)
Следовательно, решение исходного неравенства: \(x\, \in \,\left( {-\infty ;7} \right].\)
Ответ: \(\left( {-\infty ;\;7} \right].\)