已知 $\tan \alpha=\frac{1}{2}, \tan (\alpha-\beta)=-\frac{2}{5}$, 则 $\tan (\beta-2 \alpha)=$
$\because \tan (\alpha-\beta)=-\frac{2}{5}, \quad \therefore \tan (\beta-\alpha)=\frac{2}{5}$

$\tan (\beta-2 \alpha)=\tan [(\beta-\alpha)-\alpha]=\frac{\tan (\beta-\alpha)-\tan \alpha}{1+\tan (\beta-\alpha) \times \tan \alpha}=\frac{\frac{2}{5}-\frac{1}{2}}{1+\frac{2}{5} \times \frac{1}{2}}=-\frac{1}{12}$