6.001. $\frac{x^{2}+1}{x-4}-\frac{x^{2}-1}{x+3}=23$.
Solution.

Domain of definition: $x \neq-3, x \neq 4$.

$\frac{x^{2}+1}{x-4}-\frac{x^{2}-1}{x+3}=23 \Leftrightarrow \frac{16 x^{2}-25 x-275}{(x-4)(x+3)}=0 \Rightarrow$

$\Rightarrow 16 x^{2}-25 x-275=0 \Rightarrow x_{1}=-\frac{55}{16}, x_{2}=5$.

Answer: $x_{1}=-\frac{55}{16}, x_{2}=5$.