Giải: \(\sqrt {2x - 1} + \sqrt {x - 1} + 22 \\= 3x + 2\sqrt {2{x^2} - 3x + 1} \)

\(\sqrt {2x - 1}  + \sqrt {x - 1}  + 22\\= 3x + 2\sqrt {2{x^2} - 3x + 1} \)

ĐKXĐ: \(\left\{ \begin{array}{l}2x - 1 \ge 0\\x - 1 \ge 0\end{array} \right. \)

\(\Leftrightarrow \left\{ \begin{array}{l}x \ge \dfrac{1}{2}\\x \ge 1\end{array} \right. \Leftrightarrow x \ge 1\).

Đặt \(t = \sqrt {2x - 1}  + \sqrt {x - 1} \,\,\left( {t \ge 0} \right)\) ta có: \({t^2} = 2x - 1 + x - 1 + 2\sqrt {\left( {2x - 1} \right)\left( {x - 1} \right)}  \)\(= 3x - 2 + 2\sqrt {2{x^2} - 3x + 1} \)

\( \Rightarrow 3x + 2\sqrt {2{x^2} - 3x + 1}  = {t^2} + 2\)

Khi đó phương trình trở thành \(t + 22 = {t^2} + 2\)

\(\Leftrightarrow {t^2} - t - 20 = 0\)

\(\Leftrightarrow \left[ \begin{array}{l}t = 5\,\,\,\,\,\,\,\left( {tm} \right)\\t =  - 4\,\,\,\left( {ktm} \right)\end{array} \right.\)

\(\begin{array}{l}t = 5 \Rightarrow 3x + 2\sqrt {2{x^2} - 3x + 1}  = 27 \\\Leftrightarrow 2\sqrt {2{x^2} - 3x + 1}  = 27 - 3x\\ \Leftrightarrow \left\{ \begin{array}{l}27 - 3x \ge 0\\4\left( {2{x^2} - 3x + 1} \right) = 9{x^2} - 162x + 729\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}x \le 9\\{x^2} - 150x + 725 = 0\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}x \le 9\\\left[ \begin{array}{l}x = 5\\x = 145\end{array} \right.\end{array} \right.\\ \Leftrightarrow x = 5\,\,\left( {tm\,\,ĐKXĐ} \right)\end{array}\)

Vậy phương trình có nghiệm \(x = 5.\)