Question

Решите неравенство log$frac{1}{3}$ (x-2)+2 >= Log₃ (12-x)

Answer 1

$log_{ frac{1}{3}} (x-2)+2 geq log_3(12-x)$
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)