Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
a,\\
\tan x = - 2 \Leftrightarrow \dfrac{{\sin x}}{{\cos x}} = - 2 \Leftrightarrow \sin x = - 2\cos x\\
A = \dfrac{{2\sin x + 3\cos x}}{{2\cos x - 5\sin x}} = \dfrac{{2.\left( { - 2\cos x} \right) + 3\cos x}}{{2\cos x - 5.\left( { - 2\cos x} \right)}} = \dfrac{{ - \cos x}}{{12\cos x}} = - \dfrac{1}{{12}}\\
b,\\
B = \dfrac{{1 - 2{{\sin }^2}\alpha }}{{\cos \alpha + \sin \alpha }} + \dfrac{{2{{\cos }^2} - 1}}{{\cos \alpha - \sin \alpha }}\\
= \dfrac{{\left( {{{\sin }^2}\alpha + {{\cos }^2}\alpha } \right) - 2{{\sin }^2}\alpha }}{{\cos \alpha + \sin \alpha }} + \dfrac{{2{{\cos }^2}\alpha - \left( {{{\sin }^2}\alpha + {{\cos }^2}\alpha } \right)}}{{\cos \alpha - \sin \alpha }}\\
= \dfrac{{{{\cos }^2}\alpha - {{\sin }^2}\alpha }}{{\cos \alpha + \sin \alpha }} + \dfrac{{{{\cos }^2}\alpha - {{\sin }^2}\alpha }}{{\cos \alpha - \sin \alpha }}\\
= \left( {\cos \alpha - \sin \alpha } \right) + \left( {\cos \alpha + \sin \alpha } \right)\\
= 2\cos \alpha
\end{array}\)
