т вид:}

$1+\frac{11}{t-8}+\frac{28}{t^2-16t+64}\ge 0 \iff 1+\frac{11}{t-8}+\frac{28}{(t-8)^2}\ge 0 \iff$ $\iff \frac{t^2-16t+64+11t-88+28}{(t-8)^2}\ge 0 \iff \frac{t^2-5t+4}{(t-8)^2}\ge 0 \iff \frac{(t-1)(t-4)}{(t-8)^2}\ge 0.$

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

$\left[ \begin{aligned} &t\le 1,\\ &t\ne 8,\\ &t\ge 4 \end{aligned} \right. \iff \left[ \begin{aligned} &2^x\le 1,\\ &2^x\ne 8,\\ &2^x\ge 4 \end{aligned} \right. \iff \left[ \begin{aligned} &2^x\le 2^0,\\ &2^x\ne 2^3,\\ &2^x\ge 2^2 \end{aligned} \right. \iff \left[ \begin{aligned} &x\le 0,\\ &x\ne 3,\\ &x\ge 2. \end{aligned} \right.$

$\text{Таким образом, решением исходного неравенства является: } x\in(-\infty;0]\cup[2;3)\cup(3;\infty).$

$\textbf{Ответ: } (-\infty;0]\cup[2;3)\cup(3;\infty).$

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Задание №30104 ЕГЭ по Математике