Giải bất phương trình \({\log _2}\left( {x - \sqrt {{x^2} - 1} } \right).{\log _3}\left( {x + \sqrt {{x^2} - 1} } \right) = {\log _6}\left| {x - \sqrt {{x^2} - 1} } \right|\).

Điều kiện: \(\left\{ \begin{array}{l} - 1 \le x \le 1\\x - \sqrt {{x^2} - 1}  > 0\end{array} \right.\left( * \right)\)

\({\log _2}\left( {x - \sqrt {{x^2} - 1} } \right).{\log _3}\left( {x + \sqrt {{x^2} - 1} } \right) = {\log _6}\left| {x - \sqrt {{x^2} - 1} } \right|\)

\( \Leftrightarrow {\log _2}\left( {x - \sqrt {{x^2} - 1} } \right).{\log _3}\frac{1}{{x - \sqrt {{x^2} - 1} }} = {\log _6}\left( {x - \sqrt {{x^2} - 1} } \right)\)

\( \Leftrightarrow  - {\log _2}\left( {x - \sqrt {{x^2} - 1} } \right).{\log _3}6.{\log _6}\left( {x - \sqrt {{x^2} - 1} } \right) = {\log _6}\left( {x - \sqrt {{x^2} - 1} } \right)\)

\( \Leftrightarrow {\log _6}\left( {x - \sqrt {{x^2} - 1} } \right)\left[ {{{\log }_3}6.{{\log }_2}\left( {x - \sqrt {{x^2} - 1} } \right) + 1} \right] = 0\)

\( \Leftrightarrow \left[ \begin{array}{l}{\log _6}\left( {x - \sqrt {{x^2} - 1} } \right) = 0\;\left( 1 \right)\\{\log _3}6.{\log _2}\left( {x - \sqrt {{x^2} - 1} } \right) + 1 = 0\;\left( 2 \right)\end{array} \right.\)

\(\left( 1 \right) \Leftrightarrow x - \sqrt {{x^2} - 1}  = 1 \Leftrightarrow \sqrt {{x^2} - 1}  = x - 1 \Leftrightarrow \left\{ \begin{array}{l}x \ge 1\\{x^2} - 1 = {\left( {x - 1} \right)^2}\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}x \ge 1\\x = 1\end{array} \right. \Leftrightarrow x = 1\left( {tm\left( * \right)} \right)\)

\(\left( 2 \right) \Leftrightarrow {\log _3}6.{\log _2}\left( {x - \sqrt {{x^2} - 1} } \right) =  - 1 \Leftrightarrow {\log _2}\left( {x + \sqrt {{x^2} - 1} } \right) = {\log _6}3\)

\( \Leftrightarrow x + \sqrt {{x^2} - 1}  = {2^{{{\log }_6}3}} \Leftrightarrow \left\{ \begin{array}{l}x \le {2^{{{\log }_6}3}}\\{x^2} - 1 = {\left( {{2^{{{\log }_6}3}} - x} \right)^2}\end{array} \right. \Leftrightarrow x = \frac{1}{2}\left( {{2^{{{\log }_6}3}} + {2^{ - {{\log }_6}3}}} \right)\) (thỏa mãn điều kiện)