6.147. $\frac{6}{(x+1)(x+2)}+\frac{8}{(x-1)(x+4)}=1$.
## Solution.

Domain of definition: $\left\{\begin{array}{l}x \neq \pm 1, \\ x \neq-2, \\ x \neq-4 .\end{array}\right.$

Rewrite the equation as $\frac{6}{x^{2}+3 x+2}+\frac{8}{x^{2}+3 x-4}-1=0$ and let $x^{2}+3 x=y$, then we have:

$\frac{6}{y+2}+\frac{8}{y-4}-1=0 \Leftrightarrow \frac{y(y-16)}{(y+2)(y-4)}=0$, from which $\quad y_{1}=0, y_{2}=16$.

Then $x^{2}+3 x=0$ or $x^{2}+3 x=16 ; x_{1}=0, x_{2}=-3, x_{3,4}=\frac{-3 \pm \sqrt{73}}{2}$.

Answer: $x_{1}=0, x_{2}=-3, x_{3,4}=\frac{-3 \pm \sqrt{73}}{2}$.