Giải thích các bước giải:
Bài 1b:
TXĐ: \(D = R\)
Ta có:
\(\begin{array}{l}
\sqrt {{x^2} + 2017} \le \sqrt {2018} x\\
\Leftrightarrow \left\{ \begin{array}{l}
\sqrt {2018} x \ge 0\\
{\sqrt {{x^2} + 2017} ^2} \le {\left( {\sqrt {2018} x} \right)^2}
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
x \ge 0\\
{x^2} + 2017 \le 2018{x^2}
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
x \ge 0\\
2017{x^2} \ge 2017
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
x \ge 0\\
{x^2} \ge 1
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x \ge 0\\
\left[ \begin{array}{l}
x \ge 1\\
x \le - 1
\end{array} \right.
\end{array} \right. \Leftrightarrow x \ge 1
\end{array}\)
Vậy tập nghiệm của bất phương trình đã cho là \(S = \left[ {1; + \infty } \right)\)
Câu 2:
\(\begin{array}{l}
\dfrac{\pi }{2} < \alpha < \pi \Rightarrow \dfrac{\pi }{4} < \dfrac{\alpha }{2} < \dfrac{\pi }{2} \Rightarrow \left\{ \begin{array}{l}
\sin \dfrac{\alpha }{2} > 0\\
\cos \dfrac{\alpha }{2} > 0
\end{array} \right.\\
{\sin ^2}\dfrac{\alpha }{2} + {\cos ^2}\dfrac{\alpha }{2} = 1\\
\cos \dfrac{\alpha }{2} > 0 \Rightarrow \cos \dfrac{\alpha }{2} = \sqrt {1 - {{\sin }^2}\dfrac{\alpha }{2}} = \dfrac{1}{{\sqrt 5 }}\\
A = \tan \left( {\dfrac{\alpha }{2} - \dfrac{\pi }{4}} \right) = \dfrac{{\sin \left( {\dfrac{\alpha }{2} - \dfrac{\pi }{4}} \right)}}{{\cos \left( {\dfrac{\alpha }{2} - \dfrac{\pi }{4}} \right)}}\\
= \dfrac{{\sin \dfrac{\alpha }{2}.\cos \dfrac{\pi }{4} - \cos \dfrac{\alpha }{2}.sin\dfrac{\pi }{4}}}{{\cos \dfrac{\alpha }{2}.\cos \dfrac{\pi }{4} + \sin \dfrac{\alpha }{2}.sin\dfrac{\pi }{4}}}\\
= \dfrac{{\dfrac{2}{{\sqrt 5 }}.\dfrac{{\sqrt 2 }}{2} - \dfrac{1}{{\sqrt 5 }}.\dfrac{{\sqrt 2 }}{2}}}{{\dfrac{1}{{\sqrt 5 }}.\dfrac{{\sqrt 2 }}{2} + \dfrac{2}{{\sqrt 5 }}.\dfrac{{\sqrt 2 }}{2}}} = \dfrac{1}{3}
\end{array}\)
