Тип 15 № 11456

Рассмотрим три случая раскрытия модулей: $$\left[\begin{aligned} &\begin{cases} x < -1\\ 3(-x - 1) + \tfrac{1}{2}(-x + 2) - \tfrac{3}{2}x - 8 \leq 0 \end{cases}\\ &\begin{cases} -1 \leq x < 2\\ 3(x + 1) + \tfrac{1}{2}(-x + 2) - \tfrac{3}{2}x - 8 \leq 0 \end{cases}\\ &\begin{cases} x \geq 2\\ 3(x + 1) + \tfrac{1}{2}(x - 2) - \tfrac{3}{2}x - 8 \leq 0 \end{cases} \end{aligned}\right. \Leftrightarrow \left[\begin{aligned} &\begin{cases} x < -1\\ -5x - 10 \leq 0 \end{cases}\\ &\begin{cases} -1 \leq x < 2\\ x - 4 \leq 0 \end{cases}\\ &\begin{cases} x \geq 2\\ 2x - 6 \leq 0 \end{cases} \end{aligned}\right. \Leftrightarrow$$ $$\Leftrightarrow \left[\begin{aligned} &\begin{cases} x < -1\\ x \geq -2 \end{cases}\\ &\begin{cases} -1 \leq x < 2\\ x \leq 4 \end{cases}\\ &\begin{cases} x \geq 2\\ x \leq 3 \end{cases} \end{aligned}\right. \Leftrightarrow \left[\begin{aligned} &x \in \left[-2; -1\right)\\ &x \in \left[-1; 2\right)\\ &x \in \left[2; 3\right] \end{aligned}\right.$$

Ответ: $\left[-2; 3\right]$