Question

$log_{0,5}(x-3)-log_{0,5}(x+3)-log_{ frac{x+3}{x-3} } 2 textgreater 0$

Answer 1

ОДЗ x-3>0⇒x>3,x+3>0⇒x>-3,(x+3)/(x-3)>0⇒
       +        _          +
--------(-3)------(3)---------
x<-3 U X.3
x∈(3;∞)
Перейдем к основанию 2
-log(2)(x-3)+log(2)(x+3)-1/log(2)[(x+3)/(x-3)]>0
log(2)[(x+3)/(x-3)] -1/log(2)[(x+3)/(x-3)]>0
[log²(2)[(x+3)/(x-3)]-1]/log(2)[(x+3)/(x-3)]>0
(log(2)[(x+3)/(x-3)]-1)(log(2)[(x+3)/(x-3)]+1)/log(2)[(x+3)/(x-3)]>0
log(2)[(x+3)/(x-3)]=a
(a-1)(a+1)/a>0
a=1  a=-1  a=0
       _              +                _              +
 ---------(-1)----------(0)-----------(1)-----------
-1<a<0 U a>1

1){log(2)[(x+3)/(x-3)]>-1⇒(x+3)/(x-3)>1/2    (1)
   {log(2)[(x+3)/(x-3)]<0⇒(x+3)(x-3)<1        (2)
(1)  (x+3)/(x-3)-1/2>0
(2x+6-x+3)/(x-3)>0
(x+9)/(x-3)>0
x=-9  x=3
x<-9 U x>3
(2)  (x+3)/(x-3)-1<0
(x+3-x+3)/(x-3)<0
6/(x-3)<0
x-3<0
x<3
нет решения
2)log(2)[(x+3)/(x-3)]>1
(x+3)/(x-3)>2
(x+3)/(x-3)-2>0
(x+3-2x+6)/(x-3)>0
(9-x)/(x-3)>0
(x-9)/(x-3)<0
x=9  x=3
3<x<9
Ответ x∈(3;9)