\(\left( {{{\left( {\dfrac{1}{2}} \right)}^x}-16} \right)\left( {{{25}^x}-125} \right) < 0.\)
Решим неравенство методом интервалов:
\(\left( {{{\left( {\dfrac{1}{2}} \right)}^x}-16} \right)\left( {{{25}^x}-125} \right) = 0\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{{{\left( {\dfrac{1}{2}} \right)}^x}-16 = 0,}\\{{{25}^x}-125 = 0\,}\end{array}\,\,\,\,\,\,\,\, \Leftrightarrow } \right.\)
\( \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{{{\left( {\dfrac{1}{2}} \right)}^x} = {{\left( {\dfrac{1}{2}} \right)}^{-4}},}\\{{5^{2x}} = {5^3}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\end{array}\,\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{x = -4,}\\{2x = 3\,\,}\end{array}\,\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{x = -4,}\\{x = 1,5.}\end{array}} \right.} \right.} \right.\)
Следовательно, решение исходного неравенства \(x\, \in \,\left( {-\infty ;-4} \right) \cup \left( {1,5;\infty } \right).\)
Ответ: \(\left( {-\infty ;-4} \right) \cup \left( {1,5;\infty } \right).\)