Решите неравенство log (x-2)+2 >= Log₃ (12-x)

$log_{ \frac{1}{3}} (x-2)+2 \geq log_3(12-x)$
ОДЗ:
$\left \{ {{x-2\ \textgreater \ 0} \atop {12-x\ \textgreater \ 0}} \right.$
$\left \{ {{x\ \textgreater \ 2} \atop {x\ \textless \ 12}} \right.$
$x$ ∈ $(2;12)$

$log_{ 3^{-1}} (x-2)+2 \geq log_3(12-x)$
$-log_{ 3}} (x-2)+2 \geq log_3(12-x)$
$log_39 \geq log_3(12-x)+log_{ 3}} (x-2)$
$log_39 \geq log_3[(12-x)(x-2)]$
$9 \geq(12-x)(x-2)$
$9 \geq12x-24-x^2+2x$
$9 \geq-x^2+14x-24$
$x^2-14x+33 \geq 0$
$D=(-14)^2-4*1*33=64$
$x_1= \frac{14+8}{2} =11$
$x_2= \frac{14-8}{2} =3$
$(x-3)(x-11) \geq 0$

----+-------[3]----- - -----[11]-----+--------
///////////// ///////////////////
-------(2)------------------------(12)-------
//////////////////////////////

Ответ: (2;3] ∪ [11;12)