Đáp án:
$2D$
$\lim_{x \to 3}\dfrac{x^2-9}{x-3} $
$=\lim_{x \to 3} \dfrac{(x-3)(x+3)}{x-3}$
$=\lim_{x \to 3} x+3=3+3=6$
$3A$
$y=\dfrac{\sqrt[]{x-1}}{x}$
$⇒y'=\dfrac{(\sqrt[]{x-1})'.x-(\sqrt[]{x-1}).x'}{x^2}$
$=\dfrac{\dfrac{(x-1)'}{2\sqrt[]{x-1}}.x-\sqrt[]{x-1}}{x^2}$
$=\dfrac{\dfrac{x-2(x-1)}{2\sqrt[]{x-1}}}{x^2}$
$=\dfrac{x-2x+2}{2x^2\sqrt[]{x-1}}$
$=\dfrac{-x+2}{2x^2\sqrt[]{x-1}}$ $=\dfrac{2-x}{2x^2\sqrt[]{x-1}}$
BẠN THAM KHẢO.
