Phương pháp giải:

Lời giải chi tiết:

\(\begin{array}{l}a)\,\,\frac{{\sqrt {2x - 1} }}{{\sqrt {x - 1} }} = 2 \Leftrightarrow \left\{ \begin{array}{l}2x - 1 \ge 0\\x - 1 > 0\\\sqrt {2x - 1}  = 2\sqrt {x - 1} \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}x \ge \frac{1}{2}\\x > 1\\2x - 1 = 4\left( {x - 2} \right)\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}x > 1\\2x - 1 = 4x - 4\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}x > 1\\2x = 3\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}x > 1\\x = \frac{3}{2}\end{array} \right. \Leftrightarrow x = \frac{3}{2}.\end{array}\)

Vậy phương trình có nghiệm \(x = \frac{3}{2}.\)

\(\begin{array}{l}b)\,\,\frac{{\sqrt {{x^2} - 4} }}{{\sqrt {x - 2} }} = 3 \Leftrightarrow \left\{ \begin{array}{l}{x^2} - 4 \ge 0\\x - 2 > 0\\\sqrt {{x^2} - 4}  = 3\sqrt {x - 2} \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}\left[ \begin{array}{l}x \ge 2\\x \le  - 2\end{array} \right.\\x > 2\\{x^2} - 4 = 9\left( {x - 2} \right)\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}x > 2\\{x^2} - 9x + 14 = 0\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}x > 2\\\left( {x - 2} \right)\left( {x - 7} \right) = 0\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}x > 2\\\left[ \begin{array}{l}x = 2\\x = 7\end{array} \right.\end{array} \right. \Leftrightarrow x = 7.\end{array}\)

Vậy phương trình có nghiệm duy nhất \(x = 7.\)