Решите неравенство |х+1| / х(х+2)^3 меньше = 0

[latex] \frac{|x+1|}{x(x+2)^3} \leq 0[/latex] [latex]1)\ \begin{cases} x+1\ \textless \ 0 \\ \frac{x+1}{x(x+2)^3} \geq 0 \end{cases}\ = \textgreater \begin{cases} x\ \textless \ -1 \\ x(x+2)^3 \ \textless \ 0 \end{cases} =\ \textgreater \ \begin{cases} x\ \textless \ -1 \\ x(x+2) \ \textless \ 0 \end{cases} =\ \textgreater \ \\ \begin{cases} x\ \textless \ -1 \\ -2\ \textless \ x\ \textless \ 0 \end{cases} =\ \textgreater \ \ -2\ \textless \ x\ \textless \ -1.[/latex] [latex]2)\ \begin{cases} x+1 \geq 0 \\ \frac{x+1}{x(x+2)^3} \leq 0 \end{cases} =\ \textgreater \ \begin{cases} x \geq -1 \\ x(x+2)^3 \ \textless \ 0 \end{cases} =\ \textgreater \ \begin{cases} x \geq -1 \\ x(x+2) \ \textless \ 0 \end{cases} =\ \textgreater \ \\ \begin{cases} x \geq -1 \\ -2\ \textless \ x\ \textless \ 0 \end{cases} =\ \textgreater \ \ -1 \leq x\ \textless \ 0.[/latex] Объединяем решения 1) и 2) случаев: -2 < x <0. Ответ: (-2; 0).