Loq(_5) [Loq_(2,5) [ Loq_(16) (x-1/(5-x)]] > 0 ;
ОДЗ (x-1)/(5-x) >0 ⇒(x-1)/(x-5) < 0 ⇒ x ∈ (1 ; 5).
Loq_(2,5) [ Loq_(16) (x-1/(5-x)] > 5^0 ;
Loq_(2,5) [ Loq_(16) (x-1/(5-x)] > 1 ;
Loq_(16) (x-1/(5-x) >2,5;
(x-1)/(5-x)>16^2,5 ⇔(x-1)/(5-x) >1024 ;
(x-1/(5-x) -1024 >0 ;
(x-1 -5*1024 +1024x)/(5-x) >0 ;
1025(x -5121/1025)/(x-5) < 0 ⇒ x ∈(5121/5125 ;5)<br>Учитывая ОДЗ получим
ответ: (5121/5125 ; 5 ) .