Rút gọn
A=$\frac{\sqrt{3}-2sin3x}{\sqrt{3}+2sin3x}$
B=$\frac{\sqrt{2}-sinx-cosx}{sinx-cosx}$
C=$\frac{cosx.cos2x.cos4x.cos8x}{sin16x}$
Ta có

$\sqrt 2 -\sin x -\cos x = \sqrt 2 - \sqrt 2 \sin (x+\dfrac{\pi}{4})=\sqrt 2 \bigg [1-\sin (x+\dfrac{\pi}{4}) \bigg ]$

$\sin x -\cos x = \sqrt 2 \sin (x-\dfrac{\pi}{4})=-\sqrt 2 \cos (x+\dfrac{\pi}{4})$

Bạn chọn cái nào thích thì lắp vào

Câu 1

$\dfrac{\dfrac{\sqrt 3}{2} -\sin 3x}{\dfrac{\sqrt 3}{2} +\sin 3x}=\dfrac{\sin \dfrac{\pi}{3} -\sin 3x}{\sin \dfrac{\pi}{3} +\sin 3x}$

$=\dfrac{2\cos \dfrac{3x+\dfrac{\pi}{3}}{2} \sin \dfrac{\dfrac{\pi}{3}-3x}{2}}{2\sin \dfrac{3x+\dfrac{\pi}{3}}{2} \cos \dfrac{\dfrac{\pi}{3}-3x}{2}}=\cot \dfrac{3x+\dfrac{\pi}{3}}{2} . \tan \dfrac{\dfrac{\pi}{3}-3x}{2}$
$C=\frac{\cos x\cos 2x\cos 4x\cos 8x}{\sin 16x}=\frac{\cos x\cos 2x\cos 4x\cos 8x}{16\sin x\cos x\cos 2x\cos 4x\cos 8x}=\frac{1}{16\sin x}$

Bạn cần đăng nhập để có thể gửi đáp án

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