\[\begin{array}{l}
\left| {3x + 1} \right|{\rm{ }} = {\rm{ }}{x^2}{\rm{ }} + {\rm{ }}2x{\rm{ }} - {\rm{ }}3\\
+ )3{\rm{x}} + 1 \ge 0 \Leftrightarrow x \ge \frac{{ - 1}}{3}\\
= > 3{\rm{x}} + 1 = {x^2} + 2{\rm{x}} - 3\\
\Leftrightarrow {x^2} - x - 4 = 0\\
\Leftrightarrow {x^2} - 2.x.\frac{1}{2} + \frac{1}{4} - \frac{{17}}{4} = 0\\
\Leftrightarrow {\left( {x - \frac{1}{2}} \right)^2} = \frac{{17}}{4}\\
\Leftrightarrow \left[ \begin{array}{l}
x - \frac{1}{2} = \frac{{\sqrt {17} }}{2}\\
x - \frac{1}{2} = \frac{{ - \sqrt {17} }}{2}
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = \frac{{\sqrt {17} + 1}}{2}\left( {tm} \right)\\
x = \frac{{ - \sqrt {17} + 1}}{2}\left( {ktm} \right)
\end{array} \right.\\
+ )voi:x < \frac{{ - 1}}{3}\\
= > - 3{\rm{x}} - 1 = {x^2} + 2{\rm{x}} - 3\\
\Leftrightarrow {x^2} + 5{\rm{x}} - 2 = 0\\
\Leftrightarrow {x^2} + 5{\rm{x}} + \frac{{25}}{4} - \frac{{33}}{4} = 0\\
\Leftrightarrow {\left( {x + \frac{5}{2}} \right)^2} = \frac{{33}}{4}\\
\Leftrightarrow \left[ \begin{array}{l}
x + \frac{5}{2} = \frac{{\sqrt {33} }}{2}\\
x + \frac{5}{2} = \frac{{ - \sqrt {33} }}{2}
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = \frac{{\sqrt {33} - 5}}{2}\left( {ktm} \right)\\
x = \frac{{ - \sqrt {33} - 5}}{2}\left( {tm} \right)
\end{array} \right.\\
vay:S = \{ \frac{{\sqrt {17} + 1}}{2};\frac{{ - \sqrt {33} - 5}}{2}\}
\end{array}\]