Cho \(\tan \left( {2a + b + 1} \right) = 2;\tan \l...

D

- Hướng dẫn giải

\(\begin{array}{l}\,\,\,\,\,\tan \left( {2019 - 5a} \right)\\ = \frac{{\tan \left( {b - 3a + 2020} \right) - \tan \left( {2a + b + 1} \right)}}{{1 + \tan \left( {b - 3a + 2020} \right)\tan \left( {2a + b + 1} \right)}}\\ = \frac{{10 - 2}}{{1 + 10.2}} = \frac{8}{{21}}.\end{array}\)