\[\begin{array}{l}\dfrac{{3\sin \alpha - 2\cos \alpha }}{{5{{\sin }^3}\alpha + 4{{\cos }^3}\alpha }} = \dfrac{{3\tan \alpha - 2}}{{{{\cos }^2}\alpha \left[ {5{{\tan }^3}\alpha + 4} \right]}}\\ = \dfrac{{3\tan \alpha - 2}}{{5{{\tan }^3}\alpha + 4}}\left[ {1 + {{\tan }^2}\alpha } \right] = \dfrac{{70}}{{139}}\end{array}\]
Đề bài
Cho \[\tan \alpha = 3\].
Tính \[\dfrac{{2\sin \alpha + 3\cos \alpha }}{{4\sin \alpha - 5\cos \alpha }};\dfrac{{3\sin \alpha - 2\cos \alpha }}{{5{{\sin }^3}\alpha + 4{{\cos }^3}\alpha }}.\]
Lời giải chi tiết
\[\dfrac{{2\sin \alpha + 3\cos \alpha }}{{4\sin \alpha - 5\cos \alpha }} = \dfrac{{2\tan \alpha + 3}}{{4\tan \alpha - 5}} = \dfrac{9}{7}\] khi \[\tan \alpha = 3\]
\[\begin{array}{l}\dfrac{{3\sin \alpha - 2\cos \alpha }}{{5{{\sin }^3}\alpha + 4{{\cos }^3}\alpha }} = \dfrac{{3\tan \alpha - 2}}{{{{\cos }^2}\alpha \left[ {5{{\tan }^3}\alpha + 4} \right]}}\\ = \dfrac{{3\tan \alpha - 2}}{{5{{\tan }^3}\alpha + 4}}\left[ {1 + {{\tan }^2}\alpha } \right] = \dfrac{{70}}{{139}}\end{array}\]
Khi \[\tan \alpha = 3\].