Ребята, помогите, пожалуйста! Как это решается?

ОДЗ 12-x>0⇒x<12 15-x>0⇒x<15 x∈(-∞;12) log(5)(√(12-x)+1)²>-log1/(15-x) log(5)(√(12-x)+1)²/(15-x)>0 (√(12-x)+1)²/(15-x)>1 [(√(12-x)+1)²-(15-x)]/(15-x)>0 (12-x+2√(12-x)+1-15+x)/(15-x)>0 (2√(12-x)-2)/(15-x)>0 1){√(12-x)-1>0⇒√(12-x)>1⇒12-x>1⇒x<11    {15-x)>0⇒x<15 x∈(-∞;11) 2){√(12-x)-1<0⇒√(12-x)<1⇒12-x<1⇒x>11    {15-x)<0⇒x>15 нет решения Ответ x∈(-∞;11)

ОДЗ: [latex] \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][/latex] Решение: [latex]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[/latex] Ответ: [latex]x\in(-\infty;~11)[/latex]