Đáp án: $\dfrac{11}{13}$
Giải thích các bước giải:
Ta có:
$A=\dfrac{\sin^2B+2\cos^2B+1}{\sin^2B-\cos^2B+2}$
$\to A=\dfrac{\sin^2B+2\cos^2B+(\sin^2B+\cos^2B)}{\sin^2B-\cos^2B+2(\sin^2B+\cos^2B)}$
$\to A=\dfrac{2\sin^2B+3\cos^2B}{3\sin^2B+\cos^2B}$
$\to A=\dfrac{2\cdot\dfrac{\sin^2B}{\cos^2B}+3}{3\cdot \dfrac{\sin^2B}{\cos^2B}+1}$
$\to A=\dfrac{2\cdot\tan^2B+3}{3\cdot \tan^2B+1}$
$\to A=\dfrac{2\cdot 2^2+3}{3\cdot 2^2+1}$
$\to A=\dfrac{11}{13}$