\(\dfrac{{x{{\left( {x-6} \right)}^4}}}{{\left( {x-2} \right){{\left( {x + 3} \right)}^2}}} \ge 0.\)
Решим неравенство методом интервалов. Найдём нули числителя:
\(x{\left( {x-6} \right)^4} = 0\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{x = 0,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{{{\left( {x-6} \right)}^4} = 0\,\,\,\,\,\,\,}\end{array}} \right.\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{x = 0,\,}\\{x = 6.\,}\end{array}} \right.\)
Найдём нули знаменателя:
\(\left( {x-2} \right){\left( {x + 3} \right)^2} = 0\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{x-2 = 0,\,\,\,\,\,}\\{{{\left( {x + 3} \right)}^2} = 0}\end{array}} \right.\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{x = 2,\,\,\,\,\,}\\{x = -3.\,}\end{array}} \right.\)
Следовательно, решение исходного неравенства: \(x\, \in \,\left( {-\infty ;-3} \right) \cup \left( {-3;0} \right] \cup \left( {2; + \infty } \right).\)
Ответ: \(\left( {-\infty ;\;-3} \right) \cup \left( {-3;\;0} \right] \cup \left( {2;\; + \infty } \right).\)