Điều kiện xác định $x\ge -3$
$2x^2+4x=\sqrt{\dfrac{x+3}{2}}$
$\Leftrightarrow 4x^2+8x=\sqrt{2x+6}$
$\Leftrightarrow (4x^2+8x+4)+2x+2=2x+6+\sqrt{2x+6}$
$\Leftrightarrow (2x+2)^2+2x+2=2x+6+\sqrt{2x+6}$
Đặt $\left\{ \begin{array}{l} u = 2x + 2\\ v = \sqrt {2x + 6} \ge 0 \end{array} \right.$
Ta được hệ phương trình $\left\{{}\begin{matrix}u+u^2=v+v^2\\v^2-u=4\end{matrix}\right.$
$\Leftrightarrow\left\{{}\begin{matrix}\left(u-v\right)\left(u+v+1\right)=0\\v^2-u=4\end{matrix}\right.$
$u=v\Rightarrow u^2-u-4=0\Rightarrow \left[ \begin{array}{l} u = \dfrac{{1 + \sqrt {17} }}{2}\\ u = \dfrac{{1 - \sqrt {17} }}{2} \end{array} \right. \Rightarrow \left[ \begin{array}{l} x = \dfrac{{ - 3 + \sqrt {17} }}{4}\\ x = - \dfrac{{3 + \sqrt {17} }}{4} \end{array} \right. $
$v=-1-u\Rightarrow (-1-u)^2-u=4\Rightarrow u^2+u-4=0\Rightarrow \left[ \begin{array}{l} u = \dfrac{{1 + \sqrt {17} }}{2}\\ u = \dfrac{{1 - \sqrt {17} }}{2} \end{array} \right. \Rightarrow \left[ \begin{array}{l} x = \dfrac{{ - 3 + \sqrt {17} }}{4}\\ x = - \dfrac{{3 + \sqrt {17} }}{4} \end{array} \right. $
Vậy
$S = \left\{ {\frac{{ - 3 + \sqrt {17} }}{4}; - \frac{{3 + \sqrt {17} }}{4}} \right\}$
