Решите неравенство:

$log_{3}(x^2-9)-3log_{3} \frac{x+3}{x-3}\ \textgreater \ 2$
ОДЗ:
$\left \{ {{x^2-9\ \textgreater \ 0} \atop { \frac{x+3}{x-3}\ \textgreater \ 0 }} \right.$
$(x-3)(x+3)\ \textgreater \ 0$

---------+-------(-3)--------- - --------(3)--------+--------
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$x$ ∈ $(-$ ∞ $;-3)$ ∪ $(3;+$ ∞ $)$

$log_{3}[(x-3)(x+3)]-3(log_{3} (x+3)-log_3(}{x-3})) \textgreater \ 2$
$log_{3}(x-3)+log_3(x+3)-3log_{3} (x+3)+3log_3(}{x-3}) \textgreater \ 2$
$4log_{3}(x-3)-2log_{3} (x+3) \textgreater \ 2$
$2log_{3}(x-3)-log_{3} (x+3) \textgreater \ 1$
$2log_{3}(x-3)\ \textgreater \ log_{3} (x+3)+ 1$
$log_{3}(x-3)^2\ \textgreater \ log_{3} (x+3)+ log_33$
$log_{3}(x-3)^2\ \textgreater \ log_{3}[ (x+3)*3]$
$log_{3}(x-3)^2\ \textgreater \ log_{3} (3x+9)$
$(x-3)^2\ \textgreater \ 3x+9$
$x^{2} -6x+9-3x-9\ \textgreater \ 0$
$x^{2} -9x\ \textgreater \ 0$
$x(x-9)\ \textgreater \ 0$
------+------(0)------ - -------(9)-------+----------
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учтем ОДЗ

Ответ: (- ∞; -3) ∪ (9; + ∞)