Đáp án:
59 không có đáp án
60 C
Giải thích các bước giải:
$59)
M=\dfrac{\sin3x-\sin x}{\cos^2x-1}\\
=\dfrac{2\cos\dfrac{3x+x}{2}\sin\dfrac{3x-x}{2}}{\cos2x}\\
=\dfrac{2\cos2x\sin x}{\cos2x}\\
=2\sin x\\
60)
A=\dfrac{1+\cos x+\cos2x +\cos3x}{2\cos^2x+\cos x-1}\\
=\dfrac{1+\cos2x+(\cos x +\cos3x)}{2\cos^2x+\cos x-1}\\
=\dfrac{2\cos^2x+2.\cos\dfrac{x+3x}{2}\cos\dfrac{x-3x}{2}}{2\cos^2x+\cos x-1}\\
=\dfrac{2\cos^2x+2.\cos2x\cos x}{2\cos^2x+\cos x-1}\\
=\dfrac{2\cos x(\cos x+\cos2x)}{2\cos^2x+\cos x-1}\\
=\dfrac{2\cos x(2\cos^2x+\cos x-1)}{2\cos^2x+\cos x-1}\\
=2\cos x$
