Đề bài - câu 6.27 trang 199 sbt đại số 10 nâng cao

\(\begin{array}{l}\dfrac{{3\sin \alpha - 2\cos \alpha }}{{5{{\sin }^3}\alpha + 4{{\cos }^3}\alpha }} = \dfrac{{3\tan \alpha - 2}}{{{{\cos }^2}\alpha \left( {5{{\tan }^3}\alpha + 4} \right)}}\\ = \dfrac{{3\tan \alpha - 2}}{{5{{\tan }^3}\alpha + 4}}\left( {1 + {{\tan }^2}\alpha } \right) = \dfrac{{70}}{{139}}\end{array}\)

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 \(\tan \alpha = 3\).