6.023. $\frac{x+2}{x+1}+\frac{x+6}{x+3}+\frac{x+10}{x+5}=6$.
Solution.

Domain of definition: $x \neq-1, x \neq-3, x \neq-5$.

From the condition we have:

$$
\begin{aligned}
& \frac{(x+1)+1}{x+1}+\frac{(x+3)+3}{x+3}+\frac{(x+5)+5}{x+5}=6 \Leftrightarrow \\
& \Leftrightarrow \frac{x+1}{x+1}+\frac{1}{x+1}+\frac{x+3}{x+3}+\frac{3}{x+3}+\frac{x+5}{x+5}+\frac{5}{x+5}=6 \Leftrightarrow \\
& \Leftrightarrow 3+\frac{1}{x+1}+\frac{3}{x+3}+\frac{5}{x+5}=6 \Leftrightarrow \frac{3 x^{3}+18 x^{2}+23 x}{(x+1)(x+3)(x+5)}=0
\end{aligned}
$$

Considering the domain of definition, we get $3 x^{3}+18 x^{2}+23 x=0$ or $x\left(3 x^{2}+18 x+23\right)=0$, hence $x_{1}=0$, or
$3 x^{2}+18 x+23=0, x_{2}=\frac{-9+\sqrt{12}}{3}, x_{3}=\frac{-9-\sqrt{12}}{3}$.

Answer: $x_{1}=0, x_{2}=\frac{-9+\sqrt{12}}{3}, x_{3}=\frac{-9-\sqrt{12}}{3}$.