Điều kiện xác định $x\ge \dfrac 1 2$
$\begin{array}{l} \sqrt {2x - 1} - \sqrt {x + 1} = 2x - 4\\ \Leftrightarrow \dfrac{{2x - 1 - x - 1}}{{\sqrt {2x - 1} + \sqrt {x + 1} }} = 2\left( {x - 2} \right)\\ \Leftrightarrow \dfrac{{x - 2}}{{\sqrt {2x - 1} + \sqrt {x + 1} }} = 2\left( {x - 2} \right)\\ \Leftrightarrow \left( {x - 2} \right)\left( {\dfrac{1}{{\sqrt {2x - 1} + \sqrt {x + 1} }} - 2} \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l} x = 2\\ 2\left( {\sqrt {2x - 1} + \sqrt {x + 1} } \right) = 1 \end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l} x = 2\\ \sqrt {2x - 1} + \sqrt {x + 1} = \dfrac{1}{2}\left( 1 \right) \end{array} \right.\\ \sqrt {2x - 1} + \sqrt {x + 1} \ge 0 + \sqrt {1 + \dfrac{1}{2}} = \dfrac{{\sqrt 6 }}{2} > \dfrac{1}{2}\\ \Rightarrow \left( 1 \right)\left( {\text{vô nghiệm}} \right)\\ \Rightarrow S = \left\{ 2 \right\} \end{array}$
