Giải thích các bước giải:
$\begin{array}{l}
B1:\\
\left\{ \begin{array}{l}
- 6{x^2} + 7x - 1 \le 0\\
2x + 1 > 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
6{x^2} - 7x + 1 \ge 0\\
x > \dfrac{{ - 1}}{2}
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
\left( {x - 1} \right)\left( {6x - 1} \right) \ge 0\\
x > \dfrac{{ - 1}}{2}
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
\left[ \begin{array}{l}
x \ge 1\\
x \le \dfrac{1}{6}
\end{array} \right.\\
x > \dfrac{{ - 1}}{2}
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x \ge 1\\
\dfrac{{ - 1}}{2} < x \le \dfrac{1}{6}
\end{array} \right.
\end{array}$
Vậy tập nghiệm của hệ bất phương trình là: $S = \left( { - \dfrac{1}{2};\dfrac{1}{6}} \right] \cup \left[ {1; + \infty } \right)$
$\begin{array}{l}
B2:\\
a)\cos \alpha = \dfrac{2}{{\sqrt 5 }}
\end{array}$
Do $\dfrac{{ - \pi }}{2} < \alpha < 0 \Rightarrow \sin \alpha < 0$
$ \Rightarrow \sin \alpha = - \sqrt {1 - {{\cos }^2}\alpha } = \dfrac{{ - 1}}{{\sqrt 5 }}$
Khi đó:
$\begin{array}{l}
+ )\cos \left( {\alpha + \dfrac{\pi }{4}} \right)\\
= \cos \alpha .\cos \dfrac{\pi }{4} - \sin \alpha .\sin \dfrac{\pi }{4}\\
= \dfrac{2}{{\sqrt 5 }}.\dfrac{1}{{\sqrt 2 }} + \dfrac{1}{{\sqrt 5 }}.\dfrac{1}{{\sqrt 2 }}\\
= \dfrac{3}{{\sqrt {10} }}
\end{array}$
$\begin{array}{l}
+ )\sin \left( {\alpha - \dfrac{\pi }{6}} \right)\\
= \sin \alpha .\cos \dfrac{\pi }{6} - \sin \dfrac{\pi }{6}.\cos \alpha \\
= \dfrac{{ - 1}}{{\sqrt 5 }}.\dfrac{{\sqrt 3 }}{2} - \dfrac{1}{2}.\dfrac{2}{{\sqrt 5 }}\\
= \dfrac{{ - 2 - \sqrt 3 }}{{2\sqrt 5 }}\\
+ )\tan \left( {\alpha + \dfrac{\pi }{4}} \right)\\
= \dfrac{{\tan \alpha + \tan \dfrac{\pi }{4}}}{{1 - \tan \alpha .\tan \dfrac{\pi }{4}}}\\
= \dfrac{{\dfrac{{\sin \alpha }}{{\cos \alpha }} + 1}}{{1 - \dfrac{{\sin \alpha }}{{\cos \alpha }}.1}}\\
= \dfrac{{\dfrac{{ - 1}}{2} + 1}}{{1 - \left( {\dfrac{{ - 1}}{2}} \right).1}}\\
= \dfrac{1}{3}
\end{array}$
b) Ta có:
$\begin{array}{l}
\dfrac{{1 + \cos x}}{{\sin x}}\left( {1 - \dfrac{{{{\left( {1 - \cos x} \right)}^2}}}{{{{\sin }^2}x}}} \right)\\
= \dfrac{{1 + \cos x}}{{\sin x}}\left( {1 - \dfrac{{{{\left( {1 - \cos x} \right)}^2}}}{{1 - {{\cos }^2}x}}} \right)\\
= \dfrac{{1 + \cos x}}{{\sin x}}\left( {1 - \dfrac{{{{\left( {1 - \cos x} \right)}^2}}}{{\left( {1 - \cos x} \right)\left( {1 + \cos x} \right)}}} \right)\\
= \dfrac{{1 + \cos x}}{{\sin x}}\left( {1 - \dfrac{{1 - \cos x}}{{1 + \cos x}}} \right)\\
= \dfrac{{1 + \cos x}}{{\sin x}}.\dfrac{{2\cos x}}{{1 + \cos x}}\\
= \dfrac{{2\cos x}}{{\sin x}}\\
= 2\cot x
\end{array}$
