\(ĐK:x\le-\dfrac{3}{2};\dfrac{3}{2}\le x\\ Pt\Leftrightarrow\sqrt{\left(2x+3\right)\left(2x-3\right)}-2\sqrt{2x+3}=0\\ \Leftrightarrow\sqrt{2x+3}\left(\sqrt{2x-3}-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}\sqrt{2x+3}=0\\\sqrt{2x-3}=2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x+3=0\\2x-3=4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\left(tm\right)\\x=\dfrac{7}{2}\left(tm\right)\end{matrix}\right.\)
\(\sqrt{4x^2-9}=2\sqrt{2x+3}\) đk \(x\ge\dfrac{3}{2}\)
\(\Leftrightarrow4x^2-9=4\left(2x+3\right)\)
\(\Leftrightarrow4x^2-9=8x+12\)
\(\Leftrightarrow4x^2-8x-21=0\)
\(\Leftrightarrow\left(2x-7\right)\left(2x+3\right)=0\)
\(\left[{}\begin{matrix}x=\dfrac{7}{2}\left(nhận\right)\\x=-\dfrac{3}{2}\left(loại\right)\end{matrix}\right.\)
Vậy S=\(\left\{\dfrac{7}{2}\right\}\)