Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
1,\\
S = \sin 15^\circ - \cos 15^\circ = \sin 15^\circ - \cos \left( {90^\circ - 15^\circ } \right) = \sin 15^\circ - \sin 75^\circ \\
= 2.\cos \dfrac{{15^\circ + 75^\circ }}{2}.\sin \dfrac{{15^\circ - 75^\circ }}{2}\\
= 2\cos 45^\circ .\sin \left( { - 30^\circ } \right) = - \cos 45^\circ \\
= - \sin \left( {90^\circ - 45^\circ } \right) = - \sin 45^\circ = \sin \left( { - 45^\circ } \right)\\
2,\\
\left\{ \begin{array}{l}
x + 3 \ge 0\\
x + 15 \ge 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x \ge - 3\\
x \ge - 15
\end{array} \right. \Leftrightarrow x \ge - 3\\
3,\\
- \dfrac{\pi }{2} < \alpha < 0 \Rightarrow \left\{ \begin{array}{l}
\sin \alpha < 0\\
\cos \alpha > 0
\end{array} \right.\\
\sin \alpha < 0 \Rightarrow \sin \alpha = - \sqrt {1 - {{\cos }^2}\alpha } = - \dfrac{4}{5}\\
\sin \left( {\dfrac{\pi }{3} - \alpha } \right) = \sin \dfrac{\pi }{3}.\cos \alpha - \cos \dfrac{\pi }{3}.\sin \alpha = \dfrac{{\sqrt 3 }}{2}.\dfrac{3}{5} - \dfrac{1}{2}.\left( { - \dfrac{4}{5}} \right) = \dfrac{{3\sqrt 3 + 4}}{{10}}\\
4,\\
f\left( x \right) = \dfrac{1}{3}{x^2} - 4x + 13 = \dfrac{1}{3}\left( {{x^2} - 12x + 39} \right) = \dfrac{1}{3}\left[ {\left( {{x^2} - 12x + 36} \right) + 3} \right] = \dfrac{1}{3}{\left( {x - 6} \right)^2} + 1 \ge 1 > 0,\,\,\,\forall x \in R\\
5,\\
A = \dfrac{{\cos 2x + \sin 2x + {{\sin }^2}x}}{{2\sin x + \cos x}}\\
= \dfrac{{\left( {2{{\cos }^2}x - 1} \right) + 2\sin x.\cos x + \left( {1 - {{\cos }^2}x} \right)}}{{2\sin x + \cos x}}\\
= \dfrac{{{{\cos }^2}x + 2\sin x.\cos x}}{{2\sin x + \cos x}}\\
= \dfrac{{\cos x\left( {\cos x + 2\sin x} \right)}}{{2\sin x + \cos x}}\\
= \cos x\\
6,\\
\left\{ \begin{array}{l}
{x^2} - 8x + 15 \le 0\\
{x^2} - 7x + 6 \le 0\\
3x - 6 > 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
\left( {x - 3} \right)\left( {x - 5} \right) \le 0\\
\left( {x - 1} \right)\left( {x - 6} \right) \le 0\\
x > 2
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
3 \le x \le 5\\
1 \le x \le 6\\
x > 2
\end{array} \right. \Leftrightarrow 3 \le x \le 5\\
\Rightarrow S = \left[ {3;5} \right]\\
7,\\
VTCP:\,\,\,\,\overrightarrow u = \left( {\dfrac{1}{2}; - 4} \right) \Leftrightarrow \overrightarrow {{u_1}} = \left( {1; - 8} \right)
\end{array}\)
