$2.46 \frac{x-m}{x-1}+\frac{x+m}{x+1}=\frac{x-2 m}{x-2}+\frac{x+2 m}{x+2}-\frac{6(m-1)}{5}$.

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

$2.46 \frac{x-m}{x-1}+\frac{x+m}{x+1}=\frac{x-2 m}{x-2}+\frac{x+2 m}{x+2}-\frac{6(m-1)}{5}$.
2.46 After performing the transformations (do this on your own), we will obtain $-\frac{3}{5}(m-1)\left(x^{4}+4\right)=0$. It is easy to see that when $m=1$, $x$ can be any number except $\pm 1, \pm 2$; when $m \neq 1$, we get $x^{4}+4=0$, or $x^{4}=-4$ (no real roots).

Answer: if $m=1$, then $x$ is any number except $\pm 1, \pm 2$; if $m \neq 1$, there are no roots.