# 2. Solution:

Domain of definition: $x>0, 6-x>0, x \neq 1, x \neq 2 \Rightarrow x \in(0 ; 1) \cup(1 ; 2) \cup(2 ; 6)$.

We have $\frac{\left(\log _{4}(6-x)-\log _{4} 1\right)(|x-5|-|x-1|)}{\left(3^{x}-9\right)\left(3^{x}-3\right)\left(\log _{3} x-\log _{3} 1\right)} \leq 0$.

On the domain of definition, the original inequality is equivalent to the following $\frac{(5-x)(x-5-(x-1))(x-5+(x-1))}{(x-2)(x-1)^{2}} \leq 0 \Rightarrow \frac{(x-5)(x-3)}{(x-2)} \leq 0 \Rightarrow x \in(0 ; 1) \cup(1 ; 2) \cup[3 ; 5]$.
Answer: $x \in (0; 1) \cup (1; 2) \cup [3; 5]$.

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