Câu 1:
$f(2)=\dfrac{2-m}{2+1}=\dfrac{-m+2}{3}$
$\lim\limits_{x\to 2}f(x)=\lim\limits_{x\to 2}\dfrac{x^2-x-2}{x-2}=\lim\limits_{x\to 2}\dfrac{(x+1)(x-2)}{(x-2)}=\lim\limits_{x\to 2}(x+1)=2+1=3$
$\to \dfrac{-m+2}{3}=3$
$\to m=-7$
Câu 3d:
$y'=\dfrac{(5\sin x-2\cos x)'}{2\sqrt{5\sin x-2\cos x}}$
$=\dfrac{5\cos x-2.(-\sin x)}{2\sqrt{5\sin x-2\cos x}}$
$=\dfrac{2\sin x+5\cos x}{2\sqrt{5\sin x-2\cos x}}$
