Câu 6.27 trang 199 SBT Đại số 10 Nâng cao

Cho \(\tan \alpha  = 3\).

Tính \(\dfrac{{2\sin \alpha  + 3\cos \alpha }}{{4\sin \alpha  - 5\cos \alpha }};\dfrac{{3\sin \alpha  - 2\cos \alpha }}{{5{{\sin }^3}\alpha  + 4{{\cos }^3}\alpha }}.\)

Giải:

• \(\dfrac{{2\sin \alpha  + 3\cos \alpha }}{{4\sin \alpha  - 5\cos \alpha }} = \dfrac{{2\tan \alpha  + 3}}{{4\tan \alpha  - 5}} = \dfrac{9}{7}\) 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 \(\tan \alpha  = 3\).

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