\(\dfrac{{\left( {{8^x}-4} \right)\left( {{{27}^x} + 3} \right)}}{{{6^{x + 1}}-36}} > 0.\)
Решим неравенство методом интервалов.
Нули числителя:
\(\left( {{8^x}-4} \right)\left( {{{27}^x} + 3} \right) = 0\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{{8^x}-4 = 0,\,\,}\\{{{27}^x} + 3 = 0}\end{array}\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{{2^{3x}} = {2^2},}\\{{{27}^x} = -3}\end{array}\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,} \right.} \right.3x = 2\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,x = \dfrac{2}{3}.\)
Нули знаменателя:
\({6^{x + 1}}-36 = 0\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,{6^{x + 1}} = {6^2}\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,x + 1 = 2\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,x = 1.\)
Следовательно, решение исходного неравенства \(x\, \in \,\left( {-\infty ;\dfrac{2}{3}} \right) \cup \left( {1;\infty } \right).\)
Наименьшее целое положительное решение равно 2.
Ответ: 2.