a,
$\lim\limits_{x\to 1}\dfrac{f(x)-f(1)}{x-1}$
$=\lim\limits_{x\to 1}\dfrac{ \sqrt{2-x}-1}{x-1}$
$=\lim\limits_{x\to 1}\dfrac{2-x-1}{(x-1)(\sqrt{2-x}+1)}$
$=\lim\limits_{x\to 1}\dfrac{-1}{\sqrt{2-x}+1}$
$=\dfrac{-1}{\sqrt{2-1}+1}$
$=\dfrac{-1}{2}$
Vậy $f'(1)=\dfrac{-1}{2}$
b,
Với $x_o<2$:
$\lim\limits_{x\to x_o}\dfrac{f(x)-f(x_o)}{x-x_o}$
$=\lim\limits_{x\to x_o}\dfrac{ \sqrt{2-x}-\sqrt{2-x_o}}{x-x_o}$
$=\lim\limits_{x\to x_o}\dfrac{2-x-2+x_o}{(\sqrt{2-x}+\sqrt{2-x_o})(x-x_o)}$
$=\lim\limits_{x\to x_o}=\dfrac{-1}{\sqrt{2-x}+\sqrt{2-x_o}}$
$=\dfrac{-1}{2\sqrt{2-x_o}}$
Vậy trên $(-\infty;2)$, hàm số có đạo hàm $f'(x)=\dfrac{-1}{2\sqrt{2-x}}$
