Giải bất phương trình \({\log _2}\left( {x - \sqrt {{x^2} - 1} } \right).{\log _3}\left( {x + \sqrt {{x^2} - 1} } \right) = {\log _6}\left| {x - \sqrt {{x^2} - 1} } \right|\).
Nếu \(a > 0,a \ne 1\) thì \({\log _a}u\left( x \right) = {\log _a}v\left( x \right) \Leftrightarrow \left\{ \begin{array}{l}u\left( x \right) > 0\\u\left( x \right) = v\left( x \right)\end{array} \right.\) (có thể thay \(u\left( x \right) > 0\) bằng \(v\left( x \right) > 0\))
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)
Đáp án
\(x = \frac{1}{2}\left( {{2^{{{\log }_6}3}} + {2^{ - {{\log }_6}3}}} \right)\)