С полным решением Профильная математика

$\frac{6}{3-\sqrt{log_{2} (x+12)}} \geq 2 + \sqrt{log_{2}(x+12)} \\ \sqrt{log_{2}(x+12)}=t\\\\ \frac{6}{3-t} \geq 2+t \\ t \neq 3 \\ 6 \geq (2+t)(3-t) \\ 6-2t+3t-t^2 \leq 6 \\ t-t^2 \leq 0 \\ t(1-t) \leq 0 \\ \\ --0 -- 1--(3)--\ \textgreater \ t\\ - \ \ \ \ \ + \ \ \ \ \ - \\ \frac{6}{3-t}\ \textgreater \ 0\\ t\ \textless \ 3\\\\ t \in (-\infty ;0][tex]t \in (-\infty ;0] \cup [1,3) \\\\ \left \{ {{ \sqrt{log_{2}(x+12)} \leq 0\\ \atop { 1 \leq \sqrt{log_{2}(x+12)}\ \textless \ 3 \\} } \right. \\\\ ODZ \\ x\ \textgreater \ -12\\\\ 1 \leq log_{2}(x+12) \ \textless \ 9 \\ 2 \leq x+12 \ \textless \ 512 \\ -10 \leq x \ \textless \ 500$ \cup [1,3) \\ [/tex]
Так же
$log_{2}(x+12) =0 \\ x=-11$
Откуда $x \in [-11] \cup [-10,500)$