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Trả lời:
$\dfrac{\pi}{2}<\alpha<\pi⇒\cos\alpha<0$
$⇒\cos\alpha=-\sqrt{1-\sin^2\alpha}=-\dfrac{8}{17}$
$\sin\bigg{(}\dfrac{\pi}{3}-\alpha\bigg{)}=\sin\dfrac{\pi}{3}.\cos\alpha-\cos\dfrac{\pi}{3}.\sin\alpha=\dfrac{\sqrt{3}}{2}.\dfrac{-8}{17}-\dfrac{1}{2}.\dfrac{15}{17}=-\dfrac{15+8\sqrt{3}}{34}$.