6.156. $20\left(\frac{x-2}{x+1}\right)^{2}-5\left(\frac{x+2}{x-1}\right)^{2}+48 \frac{x^{2}-4}{x^{2}-1}=0$.
## Solution.

Domain of definition: $x \neq \pm 1$.

By dividing the equation by $\frac{x^{2}-4}{x^{2}-1} \neq 0$, we have

$$
\begin{aligned}
& 20 \frac{(x-2)^{2}}{(x+1)^{2}} \cdot \frac{x^{2}-1}{x^{2}-4}-5 \frac{(x+2)^{2}}{(x-1)^{2}} \cdot \frac{x^{2}-1}{x^{2}-4}+48=0 \Leftrightarrow \\
& \Leftrightarrow 20 \frac{(x-2)^{2}(x-1)(x+1)}{(x+1)^{2}(x-2)(x+2)}-5 \frac{(x+2)^{2}(x-1)(x+1)}{(x-1)^{2}(x-2)(x+2)}+48=0 \Leftrightarrow \\
& \Leftrightarrow 20 \frac{(x-2)(x-1)}{(x+2)(x+1)}-5 \frac{(x+2)(x+1)}{(x-2)(x-1)}+48=0 \Leftrightarrow
\end{aligned}
$$

$\Leftrightarrow 20 \frac{(x-2)(x-1)}{(x+2)(x+1)}-\frac{5}{\frac{(x-2)(x-1)}{(x+2)(x+1)}}+48=0$.

Let $\frac{(x-2)(x-1)}{(x+2)(x+1)}=y$, where $y \neq 0$. The equation in terms of $y$ becomes $20 y-\frac{5}{y}+48=0 \Leftrightarrow 20 y^{2}+48 y-5=0$, from which $y_{1}=-\frac{5}{2}, y_{2}=-\frac{1}{10}$.

Then

1) $\frac{(x-2)(x-1)}{(x+2)(x+1)}=-\frac{5}{2} \Leftrightarrow 7 x^{2}+9 x+14=0$, no solutions, since $D<0$;
2) $\frac{(x-2)(x-1)}{(x+2)(x+1)}=-\frac{1}{10} \Leftrightarrow 9 x^{2}-33 x+18=0$, from which $x_{3}=\frac{2}{3}, x_{4}=3$.

Answer: $x_{1}=\frac{2}{3}, x_{2}=3$.