Đáp án:
\[A = \frac{{\sqrt 6 + 3\sqrt 2 }}{2}\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
\pi < x < \frac{{3\pi }}{2} \Rightarrow \left\{ \begin{array}{l}
\sin \alpha < 0\\
\cos \alpha < 0
\end{array} \right.\\
{\sin ^2}x + {\cos ^2}x = 1\\
\sin x < 0 \Rightarrow \sin x = - \sqrt {1 - {{\cos }^2}x} = - \sqrt {1 - {{\left( { - \frac{{\sqrt 3 }}{2}} \right)}^2}} = - \frac{1}{2}\\
A = \dfrac{{\sin 2x}}{{\sin \left( {x - \frac{\pi }{4}} \right)}} = \dfrac{{2\sin x.\cos x}}{{\sin x.\cos \frac{\pi }{4} - \cos x.\sin \frac{\pi }{4}}} = \dfrac{{2.\frac{{ - 1}}{2}.\frac{{ - \sqrt 3 }}{2}}}{{\frac{{ - 1}}{2}.\frac{{\sqrt 2 }}{2} - \frac{{ - \sqrt 3 }}{2}.\frac{{\sqrt 2 }}{2}}} = \dfrac{{\sqrt 6 + 3\sqrt 2 }}{2}
\end{array}\)
Vậy \(A = \frac{{\sqrt 6 + 3\sqrt 2 }}{2}\)
