6.024. $4 x^{2}+12 x+\frac{12}{x}+\frac{4}{x^{2}}=47$.

6.024. $4 x^{2}+12 x+\frac{12}{x}+\frac{4}{x^{2}}=47$.
## Solution.

Domain of definition: $x \neq 0$.

Grouping, we get:

$$
\begin{aligned}
& \left(4 x^{2}+\frac{4}{x^{2}}\right)+\left(12 x+\frac{12}{x}\right)-47=0 \Leftrightarrow \\
& \Leftrightarrow 4\left(x^{2}+\frac{1}{x^{2}}\right)+12\left(x+\frac{1}{x}\right)-47=0
\end{aligned}
$$

Let $x+\frac{1}{x}=y \Rightarrow x^{2}+2+\frac{1}{x^{2}}=y^{2}$ or $x^{2}+\frac{1}{x^{2}}=y^{2}-2$. Then

$4\left(y^{2}-2\right)+12 y-47=0, 4 y^{2}+12 y-55=0, y_{1}=-\frac{11}{2}, y_{2}=\frac{5}{2}$.

With respect to $x$, we get two equations: $x+\frac{1}{x}=-\frac{11}{2}$ or $x+\frac{1}{x}=\frac{5}{2}$, the roots of which are $x_{1,2}=\frac{-11 \pm \sqrt{105}}{4}, x_{3}=\frac{1}{2}, x_{4}=2$.

Answer: $x_{1,2}=\frac{-11 \pm \sqrt{105}}{4}, x_{3}=\frac{1}{2}, x_{4}=2$.