Question

cho tan\(\alpha\)=3/4. tinh

A=\(\dfrac{sin^3\alpha+cos^3\alpha}{2\sin\alpha\times\cos^2\alpha+\cos\alpha\times\sin^2\alpha}\)

Answer 1

ta có : \(A=\dfrac{sin^3\alpha+cos^3\alpha}{2sin\alpha.cos^2\alpha+cos^2\alpha.sin^2\alpha}\)

\(\Leftrightarrow A=\dfrac{\dfrac{sin^3\alpha}{cos^3\alpha}+\dfrac{cos^3\alpha}{cos^3\alpha}}{\dfrac{2sin\alpha.cos^2\alpha}{cos^3\alpha}+\dfrac{cos\alpha.sin^2\alpha}{cos^3\alpha}}=\dfrac{tan^3\alpha+1}{2tan\alpha+tan^2\alpha}\)

\(\Leftrightarrow A=\dfrac{\left(\dfrac{3}{4}\right)^3+1}{2\left(\dfrac{3}{4}\right)+\left(\dfrac{3}{4}\right)^2}=\dfrac{91}{132}\)