0; \hfill \\ \end{gathered} \right. \hfill \\ \left\{ \begin{gathered} 6x - 5 = 49, \hfill \\ x > \frac{5}{6}; \hfill \\ \end{gathered} \right. \hfill \\ \left\{ \begin{gathered} x = 9, \hfill \\ x > \frac{5}{6}; \hfill \\ \end{gathered} \right. \hfill \\ \end{gathered} \]" alt="\[\begin{gathered} 2 - {\log _7}\left( {6x - 5} \right) = 0 \hfill \\ \left\{ \begin{gathered} {\log _7}\left( {6x - 5} \right) = 2, \hfill \\ 6x - 5 > 0; \hfill \\ \end{gathered} \right. \hfill \\ \left\{ \begin{gathered} 6x - 5 = 49, \hfill \\ x > \frac{5}{6}; \hfill \\ \end{gathered} \right. \hfill \\ \left\{ \begin{gathered} x = 9, \hfill \\ x > \frac{5}{6}; \hfill \\ \end{gathered} \right. \hfill \\ \end{gathered} \]" align="absmiddle" class="latex-formula">
ответ:9