Điều kiện xác định:
$\begin{array}{l} \left\{ \begin{array}{l} x - \sqrt {2x - 1} \ge 0\\ 2x - 1 \ge 0 \end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l} x \ge \dfrac{1}{2}\\ \sqrt {2x - 1} \le x \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} x \ge \dfrac{1}{2}\\ 2x - 1 \le {x^2} \end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l} x \ge \dfrac{1}{2}\\ {x^2} - 2x + 1 \ge 0 \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} x \ge \dfrac{1}{2}\\ {\left( {x - 1} \right)^2} \ge 0 \end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l} x \ge \dfrac{1}{2} \end{array} \right. \end{array}$