$%{\log _3}\left( {\frac{1}{x} - 1} \right) + {\log _3}\left( {\frac{1}{x} + 1} \right) \le {\log _3}\left( {8x - 1} \right)$%

\[\begin{array}{l} {\text{Решите неравенство:}}\\ {\log _7}\frac{3}{x} + {\log _7}\left( {{x^2} - 7x + 11} \right) \le {\log _7}\left( {{x^2} - 7x + \frac{3}{x} + 10} \right) \end{array}\]

\[2{\log _5}\left( {2x} \right) - {\log _5}\frac{x}{{1 - x}} \leqslant {\log _5}\left( {8{x^2} + \frac{1}{x} - 3} \right)\]

\[2{\log _2}\left( {x\sqrt 5 } \right) - {\log _2}\frac{x}{{1 - x}} \leqslant {\log _2}\left( {5{x^2} + \frac{1}{x} - 2} \right)\]

\[{\log _{\sqrt x }}{\left( {x - 12} \right)^2} + 1 \geqslant \log _x^2\left( {12x - {x^2}} \right)\]

\[\begin{array}{l} {\text{Решите систему неравенств}} \hfill \\ \left\{ \begin{array}{l} {\log _{3 - x}}\left( {9 - {x^2}} \right) \leqslant 1 \hfill \\ 2x + 7 + \frac{{8x + 29}}{{{x^2} + x - 6}} \geqslant - \frac{1}{{x + 3}} \hfill \\ \end{array} \right. \hfill \\ \end{array}\]