римет вид:}

$1+\frac{14}{t-9}+\frac{48}{t^2-18t+81}\ge 0 \iff 1+\frac{14}{t-9}+\frac{48}{(t-9)^2}\ge 0 \iff$ $\iff \frac{t^2-18t+81+14t-126+48}{(t-9)^2}\ge 0 \iff \frac{t^2-4t+3}{(t-9)^2}\ge 0 \iff \frac{(t-1)(t-3)}{(t-9)^2}\ge 0.$

$\text{Решим полученное неравенство методом интервалов:}$

$\left[ \begin{aligned} &t\le 1,\\ &t\ne 9,\\ &t\ge 3 \end{aligned} \right. \iff \left[ \begin{aligned} &3^x\le 1,\\ &3^x\ne 9,\\ &3^x\ge 3 \end{aligned} \right. \iff \left[ \begin{aligned} &3^x\le 3^0,\\ &3^x\ne 3^2,\\ &3^x\ge 3^1 \end{aligned} \right. \iff \left[ \begin{aligned} &x\le 0,\\ &x\ne 2,\\ &x\ge 1. \end{aligned} \right.$

$\text{Таким образом, решением исходного неравенства является: } x\in(-\infty;0]\cup[1;2)\cup(2;\infty).$

$\textbf{Ответ: } (-\infty;0]\cup[1;2)\cup(2;\infty).$

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Задание №30105 ЕГЭ по Математике