Тип 16 № 11458

Рассмотрим два случая раскрытия модуля: $$\left[\begin{aligned} &\begin{cases} 3 - 5x > 0\\ 25x^2 - 9 + 15x -30x + 9 < 0 \end{cases}\\ &\begin{cases} 3 - 5x \leq 0\\ 25x^2 - 15x + 9 - 30x + 9 < 0 \end{cases} \end{aligned}\right. \Leftrightarrow \left[\begin{aligned} &\begin{cases} x < \tfrac{3}{5}\\ 25x^2 -15x < 0 \end{cases}\\ &\begin{cases} x \geq \tfrac{3}{5}\\ 25x^2 - 45x + 18 < 0 \end{cases} \end{aligned}\right. \Leftrightarrow$$ $$\Leftrightarrow \left[\begin{aligned} &\begin{cases} x < \tfrac{3}{5}\\ x(5x - 3) < 0 \end{cases}\\ &\begin{cases} x \geq \tfrac{3}{5}\\ (5x - 3)(5x - 6) < 0 \end{cases} \end{aligned}\right. \Leftrightarrow \left[\begin{aligned} &\begin{cases} x < \tfrac{3}{5}\\ x \in (0; \tfrac{3}{5}) \end{cases}\\ &\begin{cases} x \geq \tfrac{3}{5}\\ x \in (\tfrac{3}{5}; \tfrac{6}{5}) \end{cases} \end{aligned}\right. \Leftrightarrow \left[\begin{aligned} &x \in (0; \tfrac{3}{5})\\ &x \in (\tfrac{3}{5}; \tfrac{6}{5}) \end{aligned}\right.$$

Ответ: $(0; \tfrac{3}{5}) \cup (\tfrac{3}{5}; \tfrac{6}{5})$