\(\begin{array}{l} 3{x^2} - 14\left| x \right| - 5 = 0\\ \Leftrightarrow \left[ \begin{array}{l} \left\{ \begin{array}{l} x \ge 0\\ 3{x^2} - 14x - 5 = 0 \end{array} \right.\\ \left\{ \begin{array}{l} x < 0\\ 3{x^2} + 14x - 5 = 0 \end{array} \right. \end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l} \left\{ \begin{array}{l} x \ge 0\\ \left[ \begin{array}{l} x = - \dfrac{1}{3}\left( {ktm} \right)\\ x = 5\left( {tm} \right) \end{array} \right. \end{array} \right.\\ \left\{ \begin{array}{l} x < 0\\ \left[ \begin{array}{l} x = - 5\left( {tm} \right)\\ x = \dfrac{1}{3}\left( {ktm} \right) \end{array} \right. \end{array} \right. \end{array} \right. \end{array}\)
Vậy \(S=\left\{5;-5\right\}\)
\(3x^2-14\left|x\right|-5=0\Leftrightarrow\left(\left|x\right|-5\right)\left(3\left|x\right|+1\right)\)\(=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left|x\right|=5\\\left|x\right|=-\frac{1}{3}\end{matrix}\right.\)\(\Leftrightarrow x=\pm5\)