12t+144)}{(t-9)(t+9)}\ge 0 $\Leftrightarrow \frac{2t^{2}-12t+18}{(t-9)(t+9)}\ge 0 \Leftrightarrow \frac{2(t-3)^{2}}{(t-9)(t+9)}\ge 0.$

Решая методом интервалов, получаем $t \in (-\infty;-9)\cup\{3\}\cup(9;+\infty)$ . Возвращаемся к $x$ .

$t\in(-\infty;-9)\cup\{3\}\cup(9;+\infty) \Leftrightarrow \left[ \begin{array}{l} t<-9\\ t=3\\ t>9 \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} 3^{x}<-9\\ 3^{x}=3\\ 3^{x}>9 \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} \text{таких } x \text{ нет}\\ x=1\\ x>2 \end{array} \right. .$

Ответ: $x\in\{1\}\cup(2;+\infty)$

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Задание №31134 ЕГЭ по Математике