Tính \(B = \frac{{1 + 5\sin \alpha \cos \alpha }}{{3 - 2{{\cos }},\) biết \(\tan \alpha = 2.\)

* Hướng dẫn giải

\(\begin{array}{l}B = \frac{{1 + 5\sin \alpha \cos \alpha }}{{3 - 2{{\cos }^2}\alpha }}\\ = \frac{{\frac{1}{{{{\cos }^2}\alpha }} + 5.\frac{{\sin \alpha .\cos \alpha }}{{{{\cos }^2}\alpha }}}}{{3.\frac{1}{{{{\cos }^2}\alpha }} - 2.\frac{{{{\cos }^2}\alpha }}{{{{\cos }^2}\alpha }}}}\\ = \frac{{1 + {{\tan }^2}\alpha  + 5\tan \alpha }}{{3\left( {1 + {{\tan }^2}\alpha } \right) - 2}}\\ = \frac{{1 + {2^2} + 5.2}}{{3\left( {1 + {2^2}} \right) - 2}} = \frac{{15}}{{13}}.\end{array}\)