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