6.137. $\frac{x}{x+1}+\frac{x+1}{x+2}+\frac{x+2}{x}=\frac{25}{6}$.
## Solution.

Domain of definition: $\left\{\begin{array}{l}x \neq-2, \\ x \neq-1, \\ x \neq 0 .\end{array}\right.$

Bring all terms of the equation to a common denominator:

$$
\frac{6 x^{2}(x+2)+6 x(x+1)^{2}+6(x+1)(x+2)^{2}-25 x(x+1)(x+2)}{6 x(x+1)(x+2)}=0 \Leftrightarrow
$$

$\Leftrightarrow \frac{7 x^{3}+21 x^{2}-4 x-24}{6 x(x+1)(x+2)}=0 \Leftrightarrow 7 x^{3}+21 x^{2}-4 x-24=0$ for $x \neq 0, x \neq-1$, $x \neq-2$. By checking, we find that $x_{1}=1$, since $7+21-4-24=0$. We divide the left side of the equation by $x-1: \frac{7 x^{3}+21 x^{2}-4 x-24}{x-1}=7 x^{2}+28 x+24$, therefore, the original equation can be represented as

$$
(x-1)\left(7 x^{2}+28 x+24\right)=0
$$

Solving the equation $7 x^{2}+28 x+24=0$, we find two more roots: $x_{2}=-2-\frac{2 \sqrt{7}}{7} ; x_{3}=-2+\frac{2 \sqrt{7}}{7}$.

Answer: $x_{1}=1, x_{2,3}=-2 \pm \frac{2 \sqrt{7}}{7}$.