Тип 16 № 11457

Рассмотрим два случая раскрытия модуля: $$\left[\begin{aligned} &\begin{cases} 8 - 5x > 0\\ 25x^2 - 32 + 20x - 80x + 64 < 0 \end{cases}\\ &\begin{cases} 8 - 5x \leq 0\\ 25x^2 - 20x + 32 - 80x + 64 < 0 \end{cases} \end{aligned}\right. \Leftrightarrow \left[\begin{aligned} &\begin{cases} x < \tfrac{8}{5}\\ 25x^2 - 60x + 32 < 0 \end{cases}\\ &\begin{cases} x \geq \tfrac{8}{5}\\ 25x^2 - 100x + 96 < 0 \end{cases} \end{aligned}\right. \Leftrightarrow$$ $$\Leftrightarrow \left[\begin{aligned} &\begin{cases} x < \tfrac{8}{5}\\ (5x - 4)(5x - 8) < 0 \end{cases}\\ &\begin{cases} x \geq \tfrac{8}{5}\\ (5x - 4)(5x - 12) < 0 \end{cases} \end{aligned}\right. \Leftrightarrow \left[\begin{aligned} &\begin{cases} x < \tfrac{8}{5}\\ x \in \left(\tfrac{4}{5}; \tfrac{8}{5}\right) \end{cases}\\ &\begin{cases} x \geq \tfrac{8}{5}\\ x \in \left(\tfrac{8}{5}; \tfrac{12}{5}\right) \end{cases} \end{aligned}\right. \Leftrightarrow \left[\begin{aligned} &x \in \left(\tfrac{4}{5}; \tfrac{8}{5}\right)\\ &x \in \left(\tfrac{8}{5}; \tfrac{12}{5}\right) \end{aligned}\right.$$

Ответ: $\left(\tfrac{4}{5}; \tfrac{8}{5}\right) \cup \left(\tfrac{8}{5}; \tfrac{12}{5}\right)$