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Пусть $3^{x}=t$ .

$\frac{t+9}{t-9}+\frac{t-9}{t+9}\ge \frac{12t+144}{t^{2}-81} \Leftrightarrow \frac{t+9}{t-9}+\frac{t-9}{t+9}-\frac{12t+144}{(t-9)(t+9)}\ge 0 \Leftrightarrow \frac{(t+9)(t+9)+(t-9)(t-9)-(12t+144)}{(t-9)(t+9)}\ge 0$ $\Leftrightarrow \frac{2t^{2}-12t+18}{(t-9)(t+9)}\ge 0 \Leftrightarrow \frac{2(t-3)^{2}}{(t-9)(t+9)}\ge 0.$

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: $x\in\{1\}\cup(2;+\infty)$

48:{"original_url":"https://storage.yandexcloud.net/egehub/public/776805/xROJAaiT4TVwNm1CfpuKJtAEq4ygpFHy

Задание №31134 ЕГЭ по Математике