$2.49 \frac{x^{2}+1}{x+1}+\frac{x^{2}+2}{x-2}=-2$.

Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.

$2.49 \frac{x^{2}+1}{x+1}+\frac{x^{2}+2}{x-2}=-2$.
2.49 We have

$$
\left(x^{2}+1\right)(x-2)+\left(x^{2}+2\right)(x+1)=-2(x+1)(x-2)
$$

Performing the transformations, we get $2 x^{3}+x^{2}+x-4=0$. Since the sum of all coefficients is zero, then $x=1$. Let's write the last equation in the form

$2 x^{3}-2 x^{2}+3 x^{2}-3 x+4 x-4=0$,

or $(x-1)\left(2 x^{2}+3 x+4\right)=0$.

From this, $2 x^{2}+3 x+4=0$. Since the discriminant of this equation is negative, it has no roots.

Answer: $x=1$.