Điều kiện xác định $x\ge -\dfrac 1 2$
$\begin{array}{l} 2{x^2} = \left( {3x - 2} \right)\left( {x + 1 - \sqrt {2x + 1} } \right)\\ \Leftrightarrow 2{x^2} = \left( {3x - 2} \right).\dfrac{{{x^2} + 2x + 1 - 2x - 1}}{{x + 1 + \sqrt {2x + 1} }}\\ \Leftrightarrow 2{x^2} = \left( {3x - 2} \right).\dfrac{{{x^2}}}{{x + 1 + \sqrt {2x + 1} }}\\ \Leftrightarrow {x^2}\left( {\dfrac{{3x - 2}}{{x + 1 + \sqrt {2x + 1} }} - 2} \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l} x = 0\\ 3x - 2 = 2x + 2 + 2\sqrt {2x + 1} \end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l} x = 0\\ 2\sqrt {2x + 1} = x - 4 \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} x = 0\\ 4\left( {2x + 1} \right) = {x^2} - 8x + 16 \end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l} x = 0\\ {x^2} - 16x + 12 = 0 \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} x = 0\\ x = 8 + 2\sqrt {13} \\ x = 8 - 2\sqrt {13} \end{array} \right. \end{array}$
Thử lại ta được
$S = \left\{ {0;8 + 2\sqrt {13} } \right\}$
