2. $\left|\frac{x-2}{x-1}\right|<6\left|\frac{x+1}{x+2}\right|, x \neq 1, -2$.
The first step to solve this problem is to transform the equation into
$$
\begin{array}{c}
|(x-2)(x+2)|<5|(x+1)(x-1)|, \\
x \neq 1,-2
\end{array}
$$

or $\left|x^{2}-4\right|<5\left|x^{2}-1\right|, x \neq 1,-2$. We need to find the $x$-coordinates of the intersection points of $y=4-x^{2}$ and $y=5\left(x^{2}-1\right)$, as well as the $x$-coordinates of the intersection points of $y=4-x^{2}$ and $y=5\left(1-x^{2}\right)$ (see Figure 12).
$$
\begin{aligned}
& 4-x^{2}=5 x^{2}-5 \\
9 & =6 x^{2}, \quad \frac{3}{2}=x^{2}, \\
x= & \pm \sqrt{3 / 2}
\end{aligned}
$$

and
$$
\begin{array}{c}
4-x^{2}=5-5 x^{2}, \\
4 x^{2}=1, x^{2}=\frac{1}{4}, \\
x= \pm \frac{1}{2} .
\end{array}
$$

The values that satisfy this condition form the set
$$
\left\{x \in R: x<-2 \text { or }-2<x<-\sqrt{\frac{6}{2}}\right.
$$

or $-\frac{1}{2}<x<\frac{1}{2}$ or $\left.\sqrt{\frac{6}{2}}<x\right\}$.