5. Solve the inequality

$$
\left(\log _{\frac{3}{2} x^{2}-\frac{4}{3} x+\frac{5}{6}}\left(1+4 x^{2}\right) \cdot \log _{\frac{3}{2} x^{2}-\frac{4}{3} x+\frac{5}{6}}\left(1-4 x^{2}\right)+1\right) \log _{1-16 x^{4}}\left(\frac{3 x^{2}}{2}-\frac{4 x}{3}+\frac{5}{6}\right) \geqslant 1
$$
Answer: $x \in\left(-\frac{1}{2} ;-\frac{1}{3}\right] \cup\left[-\frac{1}{5} ;-\frac{1}{9}\right) \cup\left(-\frac{1}{11} ; 0\right) \cup\left(0 ; \frac{1}{3}\right]$.

Solution. The domain of definition of the logarithms in the inequality is determined by the conditions

$$
\begin{gathered}
1-4 x^{2}>0 \Leftrightarrow-\frac{1}{2}0 \Leftrightarrow x \in \mathbb{R}, \\
1+4 x^{2} \neq 1 \Leftrightarrow x \neq 0 \\
\frac{3 x^{2}}{2}-\frac{4 x}{3}+\frac{5}{6}>0 \Leftrightarrow x \in \mathbb{R}, \\
\frac{3 x^{2}}{2}-\frac{4 x}{3}+\frac{5}{6} \neq 1 \Leftrightarrow\left\{\begin{array}{l}
x \neq-\frac{1}{9} \\
x \neq 1
\end{array}\right.
\end{gathered}
$$

In the end, we get $x \in\left(-\frac{1}{2} ; \frac{1}{2}\right)$ and $x \neq 0, x \neq-\frac{1}{9}$.

Let $\frac{3 x^{2}}{2}-\frac{4 x}{3}+\frac{5}{6}=u, 1+4 x^{2}=v, 1-4 x^{2}=w$. We write and transform the inequality:

$$
\begin{aligned}
&\left(\log _{u} v \cdot \log _{u} w+1\right) \log _{v w} u \geqslant 1 \Leftrightarrow \frac{\log _{u} v \cdot \log _{u} w+1}{\log _{u} v w}-1 \geqslant 0 \Leftrightarrow \\
& \Leftrightarrow \frac{\log _{u} v \cdot \log _{u} w+1-\log _{u} v-\log _{u} w}{\log _{u} v w} \geqslant 1 \Leftrightarrow \frac{\left(\log _{u} v-1\right)\left(\log _{u} w-1\right)}{\log _{u} v w} \geqslant 0
\end{aligned}
$$

To solve this inequality, we further apply the method of rationalization: the sign of the difference $\log _{a} b- \log _{a} c$ on the domain of admissible values coincides with the sign of the expression $\frac{b-c}{a-1}$; in particular (when $c=1$), the sign of the logarithm $\log _{a} b$ coincides with the sign of the expression $\frac{b-1}{a-1}$. Then from the last inequality we get $\frac{(u-1)(v-u)(u-1)(w-u)}{(u-1)(v w-1)} \geqslant 0 \Leftrightarrow \frac{(v-u)(w-u)}{(u-1)(v w-1)} \geqslant 0$. Substituting the expressions for $u, v, w$ here and solving the resulting inequality:

$$
\begin{aligned}
& \frac{\left(\frac{5 x^{2}}{2}+\frac{4 x}{3}+\frac{1}{6}\right)\left(-\frac{11 x^{2}}{2}+\frac{4 x}{3}+\frac{1}{6}\right)}{\left(\frac{3 x^{2}}{2}-\frac{4 x}{3}-\frac{1}{6}\right)\left(-16 x^{4}\right)} \geqslant 0 \Leftrightarrow \frac{\left(15 x^{2}+8 x+1\right)\left(33 x^{2}-8 x-1\right)}{x^{2}\left(9 x^{2}-8 x-1\right)} \geqslant 0 \Leftrightarrow \\
& \Leftrightarrow \frac{(5 x+1)(3 x+1)(3 x-1)(11 x+1)}{x^{2}(x-1)(9 x+1)} \geqslant 0 \Leftrightarrow x \in\left(-\infty ;-\frac{1}{3}\right] \cup\left[-\frac{1}{5} ;-\frac{1}{9}\right) \cup\left[-\frac{1}{11} ; 0\right) \cup\left(0 ; \frac{1}{3}\right] \cup(1 ;+\infty)
\end{aligned}
$$

Taking into account the domain of definition, we are left with $x \in\left(-\frac{1}{2} ;-\frac{1}{3}\right] \cup\left[-\frac{1}{5} ;-\frac{1}{9}\right) \cup\left(-\frac{1}{11} ; 0\right) \cup\left(0 ; \frac{1}{3}\right]$.