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Trả lời:
$a,\dfrac{3\pi}{2}<a<2\pi⇒\begin{cases}\sin a<0\\\cos a>0\end{cases}$
$\cos a=\dfrac{4}{5}$
$⇒\sin a=-\sqrt{1-\cos^2a}=-\sqrt{1-\bigg{(}\dfrac{4}{5}\bigg{)}^2}=-\dfrac{3}{5}$
$b,\sin a-\cos a=\dfrac{5}{4}$
$⇒\sin^2a-2\sin a\cos a+\cos^2a=\dfrac{25}{16}$
$⇒1-2\sin a\cos a=\dfrac{25}{16}$
$⇒\sin a\cos a=-\dfrac{9}{32}$
$A=\sin^3a-\cos^3a$
$=(\sin a-\cos a)(\sin^2a+\sin a\cos a+\cos^2a)$
$=\dfrac{5}{4}.\bigg{(}1-\dfrac{9}{32}\bigg{)}$
$=\dfrac{115}{128}$.
