5. Solve the inequality

$$
\frac{\left(2 \cdot 3^{-\log _{x} 2}-6\right) \sqrt{2-\sqrt{2 \log _{x} 2+3}}}{2+\sqrt{\log _{x} 2+2}}>\frac{\left(3^{-\log _{x} 2}-3\right) \sqrt{2-\sqrt{2 \log _{x} 2+3}}}{\sqrt{\log _{x} 2+2}-1}
$$
Solution: Let's make the substitution $y=\log _{x} 2$.

$$
\begin{aligned}
& \frac{\left(2 \cdot 3^{-y}-6\right) \sqrt{2-\sqrt{2 y+3}}}{2+\sqrt{y+2}}>\frac{\left(3^{-y}-3\right) \sqrt{2-\sqrt{2 y+3}}}{\sqrt{y+2}-1} \Leftrightarrow \\
& \left(3^{-y}-3\right)\left(\frac{2}{2+\sqrt{y+2}}-\frac{1}{\sqrt{y+2}-1}\right) \sqrt{2-\sqrt{2 y+3}}>0
\end{aligned}
$$

Find the domain: $y \geq-2, \quad y \neq-1,2-\sqrt{2 y+3} \geq 0 \Rightarrow-1.5 \leq y \leq 0.5$. Since the inequality is strict, then for $-1.5 \leq y < 0$, or $(y+1)\left(\frac{\sqrt{y+2}-4}{(2+\sqrt{y+2})(\sqrt{y+2}-1)}\right) > 0 , } \\
{ \log _{2} x \leq -2 / 3 , } \\
{ \left[ \begin{array} { c } 
{ \log _{2} x < 2 , } \\
{ x \neq 1 / 2 , }
\end{array} \right]
\end{array} \Leftrightarrow \left\{\begin{array}{c}
x \in(0 ; 1 / \sqrt[3]{4}] \cup(1 ;+\infty), \\
x \in(0 ; 1) \cup(4 ;+\infty), \\
x \neq 1 / 2,
\end{array} \Leftrightarrow\right.\right.\right. \\
& x \in(0 ; 1 / 2) \cup(1 / 2 ; 1 / \sqrt[3]{4}] \cup(4 ;+\infty) .
\end{aligned}
$$

Answer: $x \in(0 ; 1 / 2) \cup(1 / 2 ; 1 / \sqrt[3]{4}] \cup(4 ;+\infty)$.