Тип 16 № 11224

$$\frac{8^{x + 1} - 40}{2\cdot64^x - 32} \leq 1 \Leftrightarrow \frac{8\cdot8^x - 40}{2\cdot8^{2x} - 32} \leq 1$$

Пусть $8^x = t > 0$

Имеем: $$\frac{8t - 40}{2t^2 - 32} \leq 1$$ $$\frac{8t - 40}{2t^2 - 32} - 1 \leq 0$$ $$\frac{8t - 40 -2t^2 + 32}{2t^2 - 32} \leq 0$$ $$\frac{-2t^2 + 8t - 8}{2t^2 - 32} \leq 0$$ $$\frac{-t^2 + 4t - 4}{t^2 - 16} \leq 0 \Leftrightarrow \frac{t^2 - 4t + 4}{t^2 - 16}\geq 0$$ $$\frac{(t - 2)^2}{(t - 4)(t + 4)}\geq 0$$

Решение данного неравенства равносильно совокупности:

$$\left[\begin{aligned} &t > 4 \\ &t = 2\\ \end{aligned}\right. \Leftrightarrow \left[\begin{aligned} &8^x > 4 \\ &8^x = 2\\ \end{aligned}\right. \Leftrightarrow \left[\begin{aligned} &2^{3x} > 2^2 \\ &2^{3x} = 2\\ \end{aligned}\right.\Leftrightarrow\left[\begin{aligned} &x > \tfrac{2}{3} \\ &x = \tfrac{1}{3}\\ \end{aligned}\right.$$

Ответ: $\{\tfrac{1}{3}\} \cup (\tfrac{2}{3}; +\infty)$