Cho \(tan\alpha=2,\) tính \(A=\frac{2sin\alpha-cos\alpha}{3sin\alpha+2cos\alpha}\).
Hướng dẫn giải
Ta có \(cos\alpha\ne0,\) nên \(A=\frac{2sin\alpha-cos\alpha}{3sin\alpha+2cos\alpha}=\frac{\frac{2sin\alpha}{cos\alpha}-1}{\frac{3sin\alpha}{cos\alpha}+2}=\frac{2tan\alpha-1}{3tan\alpha+2}\)
\(=\frac{2.2-1}{3.2+2}=\frac{3}{8}.\)