Giải hệ bất phương trình cho sau: \(\left\{ \begin{array}{l}\dfrac{{2 - 3x}}{{4x - 1}} \le 0\\{\left( {x + 1} \right)^2} - 16 > 0\end{array} \right.\).

 \(\left\{ \begin{array}{l}\dfrac{{2 - 3x}}{{4x - 1}} \le 0\\{\left( {x + 1} \right)^2} - 16 > 0\end{array} \right.\)  

ĐKXĐ: \(4x - 1 \ne 0 \Leftrightarrow x \ne \dfrac{1}{4}\)

\( \Leftrightarrow \left\{ \begin{array}{l}\dfrac{{2 - 3x}}{{4x - 1}} \le 0\\{x^2} + 2x + 1 - 16 > 0\end{array} \right.\)\( \Leftrightarrow \left\{ \begin{array}{l}\dfrac{{2 - 3x}}{{4x - 1}} \le 0\\{x^2} + 2x - 15 > 0\end{array} \right.\)

\( \Leftrightarrow \left\{ \begin{array}{l}\left[ \begin{array}{l}\left\{ \begin{array}{l}2 - 3x \ge 0\\4x - 1 < 0\end{array} \right.\\\left\{ \begin{array}{l}2 - 3x \le 0\\4x - 1 > 0\end{array} \right.\end{array} \right.\\\left[ \begin{array}{l}x <  - 5\\x > 3\end{array} \right.\end{array} \right.\)\( \Leftrightarrow \left\{ \begin{array}{l}\left[ \begin{array}{l}x < \dfrac{1}{4}\\x \ge \dfrac{2}{3}\end{array} \right.\\\left[ \begin{array}{l}x <  - 5\\x > 3\end{array} \right.\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x <  - 5\\x > 3\end{array} \right.\)

Vậy hệ bất phương trình có tập nghiệm \(S = \left( { - \infty ;\,\, - 5} \right) \cup \left( {3;\,\, + \infty } \right)\).