[tex]log_{0,5}(x-3)-log_{0,5}(x+3)-log_{ \frac{x+3}{x-3} } 2\ \textgreater \ 0[/tex]

ОДЗ x-3>0⇒x>3,x+3>0⇒x>-3,(x+3)/(x-3)>0⇒       +        _          +--------(-3)------(3)---------x<-3 U X.3x∈(3;∞)Перейдем к основанию 2-log(2)(x-3)+log(2)(x+3)-1/log(2)[(x+3)/(x-3)]>0log(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)]>0log(2)[(x+3)/(x-3)]=a(a-1)(a+1)/a>0a=1  a=-1  a=0       _              +                _              + ---------(-1)----------(0)-----------(1)------------1<a<0 U a>11){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)>0x=-9  x=3x<-9 U x>3(2)  (x+3)/(x-3)-1<0(x+3-x+3)/(x-3)<06/(x-3)<0x-3<0x<3нет решения2)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)<0x=9  x=33<x<9Ответ x∈(3;9)