Điều kiện xác định: $ - \dfrac{5}{2} \le x \le 3$
$\begin{array}{l}
\sqrt {2x + 5} - \sqrt {3 - x} = 2\\
\Leftrightarrow \sqrt {2x + 5} = 2 + \sqrt {3 - x} \\
\Leftrightarrow 2x + 5 = 4 + \left( {3 - x} \right) + 4\sqrt {3 - x} \\
\Leftrightarrow 2x + 5 = 7 - x + 4\sqrt {3 - x} \\
\Leftrightarrow 3x - 2 = 4\sqrt {3 - x} \\
\Leftrightarrow {\left( {3x - 2} \right)^2} = 16\left( {3 - x} \right)\\
\Leftrightarrow 9{x^2} - 12x + 4 = 48 - 16x\\
\Leftrightarrow 9{x^2} + 4x - 44 = 0\\
\Leftrightarrow 9{x^2} - 18x + 22x - 44 = 0\\
\Leftrightarrow 9x\left( {x - 2} \right) + 22\left( {x - 2} \right) = 0\\
\Leftrightarrow \left( {x - 2} \right)\left( {9x + 22} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x = 2(tm)\\
x = - \dfrac{{22}}{9}(L)
\end{array} \right.\\
\Rightarrow S = \left\{ 2 \right\}
\end{array}$
