Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
\dfrac{{2{{\cos }^2}x + \cos x - 1}}{{1 + \cos x + \cos 2x + \cos 3x}}\\
= \dfrac{{\left( {2{{\cos }^2}x - 1} \right) + \cos x}}{{\left( {1 + \cos 2x} \right) + \left( {\cos 3x + \cos x} \right)}}\\
= \dfrac{{\cos 2x + \cos x}}{{\left( {1 + \left( {2{{\cos }^2}x - 1} \right)} \right) + 2.\cos \dfrac{{3x + x}}{2}.\cos \dfrac{{3x - x}}{2}}}\\
= \dfrac{{\cos 2x + \cos x}}{{2{{\cos }^2}x + 2\cos 2x.\cos x}}\\
= \dfrac{{\cos 2x + \cos x}}{{2\cos x.\left( {\cos x + \cos 2x} \right)}}\\
= \dfrac{1}{{2\cos x}}
\end{array}\)
