Question

Решите систему неравенств
$\left \{ {{\frac{1}{(x+4)}-\frac{1}{(2x+4)}\ \textless \ =0} \atop {x^2+5x+4\ \textless \ =0} \right.$

Answer 1

$\left\{ \begin{gathered} \frac{1}{x+4} - \dfrac{1}{2x+4} \le 0 \\ x^{2} + 5x + 4 \le 0 \\ \end{gathered} \right.$

$ODZ: \ \left\{ \begin{gathered} x + 4 \ne 0 \\ 2x + 4 \ne 0 \\ \end{gathered} \right. \ ; \ \left\{ \begin{gathered} x \ne -4 \\ x \ne -2 \\ \end{gathered} \right.$

$\left\{ \begin{gathered} \ \frac{2x+4 - x - 4}{(2x+4)(x+4)} \le 0 \\ x^{2} + 5x + 4 \le 0 \\ \end{gathered} \right.$

$\left\{ \begin{gathered} \ \frac{x}{2(x+2)(x+4)} \le 0\ (1) \\ (x+1)(x+4) \le 0\ (2) \\ \end{gathered} \right.$

$\left\{ \begin{gathered} x \in (-\infty ; -4)\cup (-2;0] \\ x \in [-4;-1] \\ \end{gathered} \right.$

Пересечем множество решений и учтём ОДЗ (3):

$x \in (-2;-1]$

Ответ: x ∈ (-2;-1]