Помогите. пожалуйста, очень нужно!

ОДЗ:

$\left \{ {{\big{ \sqrt{12-x}+1\ \textgreater \ 0} } \atop { \frac{\big1}{\big{15-x}}\ \textgreater \ \big{0} }} \right. ~~\left \{ {{ \big{\sqrt{12-x}\ \textgreater \ -1 } \atop {\big{15-x\ \textgreater \ 0}} \right. ~~~~ \left \{ {{ \big{\sqrt{12-x} \geq 0} \atop {\big{x\ \textless \ 15}} \right. \\ \\ \\ \left \{ {{\big{12-x \geq 0} \atop {\big{x\ \textless \ 15}} \right. ~~~~ \left \{ {{\big{x \leq 12} \atop {\big{x\ \textless \ 15}} \right. ~~\Rightarrow~x\in(-\infty;12]$

Решение:

$2log_5( \sqrt{12-x}+1 )\ \textgreater \ log_{ \frac{1}{5} }( \frac{1}{15-x} ) \\ log_5( \sqrt{12-x}+1 )^2\ \textgreater \ -log_5( \frac{1}{15-x} ) \\ log_5( \sqrt{12-x}+1 )^2\ \textgreater \ log_5(15-x) \\ ( \sqrt{12-x}+1 )^2\ \textgreater \ 15-x \\ 12-x+2\sqrt{12-x}+1\ \textgreater \ 15-x \\2\sqrt{12-x}-x+13\ \textgreater \ 15-x \\ 2\sqrt{12-x}\ \textgreater \ 15-x+x-13 \\ 2\sqrt{12-x}\ \textgreater \ 2|:2 \\ \sqrt{12-x} \ \textgreater \ 1 \\ 12-x\ \textgreater \ 1 \\ x\ \textless \ 11$

Ответ: $x\in(-\infty;~11)$