Question

[Le Phuong Thanh]

Cho \(\tan\alpha\) =2. Giá trị biểu thức : \(\frac{4.\sin^2\alpha+3.\cos\alpha.\sin\alpha}{5.\sin^2\alpha-2.\cos^2\alpha}\)=?

Answer 1

[Nguyen]

Đặt bthức =A.

Ta có:\(A=\frac{4\frac{\sin^2\alpha}{\cos^2\alpha}+3.\frac{\cos\alpha}{\sin\alpha}}{5.\frac{\sin^2\alpha}{\cos^2\alpha}-2}\)\(=\frac{4\tan^2\alpha+3\tan\alpha}{5\tan^2\alpha-2}\)\(=\frac{11}{9}\)