Đáp án:
$A=\dfrac{11}{14}$
Giải thích các bước giải:
$A=\dfrac{\sin^2x-\sin x\cos x+2\cos^2x}{\sin x\cos x+4\cos^2x}\\
=\dfrac{\dfrac{\sin^2x}{\sin^2x}-\dfrac{\sin x\cos x}{\sin^2x}+\dfrac{2\cos^2x}{\sin^2x}}{\dfrac{\sin x\cos x}{\sin^2x}+\dfrac{4\cos^2x}{\sin^2x}}\\
=\dfrac{1-\dfrac{\cos x}{\sin x}+\dfrac{2\cos^2x}{\sin^2x}}{\dfrac{\cos x}{\sin x}+\dfrac{4\cos^2x}{\sin^2x}}\\
=\dfrac{1-\cot x+2\cot^2x}{\cot x+4\cot^2x}\\
=\dfrac{1-(-2)+2.(-2)^2}{-2+4.(-2)^2}\\
=\dfrac{1+2+8}{-2+16}\\
=\dfrac{11}{14}$
