\[\begin{array}{l} {\text{Решите неравенство:}}\\ x + \frac{{20}}{{x + 6}} \ge 6 \end{array}\]

$$\eqalign{ {\text{Решите неравенство: 1}}{{\text{6}}^{x - 1}} - 67 \cdot {4^{x - 2}} + 12 \le 0. } $$

\[{\text{Решите неравенство:}}\] $$\eqalign{ \frac{1}{{{3^x} - 1}} + \frac{{{9^{x + \frac{1}{2}}} - {3^{x + 3}} + 3}}{{{3^x} - 9}} \ge {3^{x + 1}}. } $$

\[{\text{Решите неравенство:}}\] $$\frac{{{2^x}}}{{{2^x} - 3}} + \frac{{{2^x} + 1}}{{{2^x} - 2}} + \frac{5}{{{4^x} - 5 \cdot {2^x} + 6}} \leqslant 0$$

\[\frac{{{{\left( {\sqrt 5 + 2} \right)}^x} - 2\sqrt 5 + {{\left( {\sqrt 5 - 2} \right)}^x}}}{{1 - {{\left( {\sqrt 5 - 2} \right)}^x}}} > 0\]

\[\begin{array}{l} {\text{Решите неравенство:}} \hfill \\ {7^{2{x^2} - 8x + 7}} - 10 \cdot {14^{{x^2} - 4x + 3}} + {3^{2{x^2} - 8x + 7}} \geqslant 0. \hfill \\ \end{array}\]

$$\eqalign{ {\text{Решите неравенство:}} } $$ $$\eqalign{ \frac{2}{{{{\log }_2}x}} + \frac{5}{{\log _2^2x - {{\log }_2}{x^3}}} \le \frac{{{{\log }_2}x}}{{{{\log }_2}\left( {\frac{x}{8}} \right)}}. } $$

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

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

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

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

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

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