6.136. $\frac{x^{2}+1}{x+1}+\frac{x^{2}+2}{x-2}=-2$.

6.136. $\frac{x^{2}+1}{x+1}+\frac{x^{2}+2}{x-2}=-2$. 

(Note: The equation is the same in both languages, so the translation is identical to the original text.)
## Solution.

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

After bringing all terms of the equation to a common denominator, we have

$$
\begin{aligned}
& \frac{\left(x^{2}+1\right)(x-2)+\left(x^{2}+2\right)(x+1)+2(x+1)(x-2)}{(x+1)(x-2)}=0 \Leftrightarrow \\
& \Leftrightarrow \frac{2 x^{3}+x^{2}+x-4}{(x+1)(x-2)}=0 \Leftrightarrow 2 x^{3}+x^{2}+x-4=0 \text { for } x \neq-1 \text { and } x \neq 2
\end{aligned}
$$

Rewrite this equation as $\left(2 x^{3}-2\right)+\left(x^{2}-1\right)+(x-1)=0$, $2(x-1)\left(x^{2}+x+1\right)+(x-1)(x+1)+x-1=0 \Leftrightarrow(x-1)\left(2 x^{2}+3 x+4\right)=0$, from which $x-1=0$, or $2 x^{2}+3 x+4=0 ; x_{1}=1$, for the quadratic equation $D<0$.

Answer: $x=1$.