# 2. Solution:

Domain of definition: $x+6 \geq 0, x+1>0,5-x>0, x \neq 0, x \neq 1, x \neq 4 \Rightarrow x \in(-1 ; 0) \cup(0 ; 1) \cup(1 ; 4) \cup(4 ; 5)$.

We have $\frac{(\sqrt{x+6}-1)(\sqrt{x+6}-3)\left(\log _{2}(x+1)-\log _{2} 1\right)}{\left(2^{x}-1\right)\left(2^{x}-2\right)\left(\log _{5}(5-x)-\log _{5} 1\right)} \geq 0$.

On the domain of definition, the original inequality is equivalent to the following

$$
\frac{(x+5)(x-3) x}{x(x-1)(4-x)} \geq 0 \Rightarrow \frac{(x-3)}{(x-1)(4-x)} \geq 0 \Rightarrow x \in(-1 ; 0) \cup(0 ; 1) \cup[3 ; 4)
$$
Answer: $x \in(-1 ; 0) \cup(0 ; 1) \cup[3 ; 4)$.