Đáp án:
$A=±\dfrac{\sqrt{7}}{4}$
Giải thích các bước giải:
$\sin\alpha+\cos\alpha=\dfrac{5}{4}$
$⇒(\sin\alpha+\cos\alpha)^2=\dfrac{25}{16}$
$⇒\sin^2\alpha+2\sin\alpha\cos\alpha+\cos^2\alpha=\dfrac{25}{16}$
$⇒1+2\sin\alpha\cos\alpha=\dfrac{25}{16}$
$⇒ 2\sin\alpha\cos\alpha=\dfrac{9}{16}$
$A=\sin\alpha-\cos\alpha$
$A^2=(\sin\alpha-\cos\alpha)^2$
$=\sin^2\alpha-2\sin\alpha\cos\alpha+\cos^2\alpha$
$=1-2\sin\alpha\cos\alpha$
$=1-\dfrac{9}{16}=\dfrac{7}{16}$
$⇒A=±\dfrac{\sqrt{7}}{4}$.
