10. Given the sequence $\left\{x_{n}\right\}, x_{1}=1$, and $x_{n+1}=\frac{\sqrt{3} x_{n}+1}{\sqrt{3}-x_{n}}, x_{1999}-x_{601}=$
10. 0 .
$x_{n+1}=\frac{x_{n}+\frac{\sqrt{3}}{3}}{1-\frac{\sqrt{3}}{3} x_{n}}$. Let $x_{n}=\tan \alpha_{n}$, then $x_{n+1}=\tan \alpha_{n+1}=\tan \left(\alpha_{n}+\frac{\pi}{6}\right)$, so $x_{n+6}=x_{n}$, which means $\left\{x_{n}\right\}$ is a periodic sequence. $1999=333 \times 6+1,601=6 \times 100+1$, thus $x_{1999}-x_{601}=0$.