Решите неравенство log[latex] \frac{1}{3} [/latex] (x-2)+2 больше = Log₃ (12-x)

[latex]log_{ \frac{1}{3}} (x-2)+2 \geq log_3(12-x)[/latex] ОДЗ:  [latex] \left \{ {{x-2\ \textgreater \ 0} \atop {12-x\ \textgreater \ 0}} \right. [/latex] [latex] \left \{ {{x\ \textgreater \ 2} \atop {x\ \textless \ 12}} \right. [/latex] [latex]x[/latex] ∈ [latex](2;12)[/latex] [latex]log_{ 3^{-1}} (x-2)+2 \geq log_3(12-x)[/latex] [latex]-log_{ 3}} (x-2)+2 \geq log_3(12-x)[/latex] [latex]log_39 \geq log_3(12-x)+log_{ 3}} (x-2)[/latex] [latex]log_39 \geq log_3[(12-x)(x-2)][/latex] [latex]9 \geq(12-x)(x-2)[/latex] [latex]9 \geq12x-24-x^2+2x[/latex] [latex]9 \geq-x^2+14x-24[/latex] [latex]x^2-14x+33 \geq 0[/latex] [latex]D=(-14)^2-4*1*33=64[/latex] [latex]x_1= \frac{14+8}{2} =11[/latex] [latex]x_2= \frac{14-8}{2} =3[/latex] [latex](x-3)(x-11) \geq 0[/latex] ----+-------[3]----- - -----[11]-----+-------- /////////////                      /////////////////// -------(2)------------------------(12)-------             ////////////////////////////// Ответ: (2;3] ∪ [11;12)