ath/tex; mode=display">\begin{aligned}\left(\left(t^2-5t\right)-24\right)\left(\left(t^2-5t\right)+4\right)\leq0\\\left(t^2-5t-24\right)\left(t^2-5t+4\right)\leq0\end{aligned}

Разложим на множители

$(t+3)(t-8)(t-1)(t-4)\leq0$

Отсюда по методу интервалов получаем

$\begin{aligned}t\in[-3;1]\cup[4;8]\\\text{При }-3\leq t\leq1\text{ получим }-3\leq2^x\leq1,\text{откуда }x\leq0\\\text{При }4\leq t\leq8\text{ получим }4\leq2^x\leq8,\text{откуда }2\leq x\leq3\end{aligned}$

Объединяя, получаем решения исходного неравенства:

$x\leq0;2\leq x\leq3$

$\text{Ответ: }(-\infty;0]\cup[2;3]$

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Задание №3834 ЕГЭ по Математике