Ta có
$\underset{x \to 2^-}{\lim} f(x) = \underset{x \to 2^-}{\lim} \dfrac{|2x^2 - 7x + 6|}{x-2}$
$= \underset{x \to 2^-}{\lim} \dfrac{|(x-2)(2x-3)|}{x-2}$
$= \underset{x \to 2^-}{\lim} \pm 1 |2x-3|
$= \pm 1$
Khi $x \to 2$ thì $2x^2 - 7x + 6 < 0$. Vậy $|2x^2 - 7x + 6| = -2x^2 + 7x - 6$
Vậy
$\underset{x \to 2^-}{\lim} f(x) = -1$
Lại có
$\underset{x \to 2^+}{\lim} f(x) = \underset{x \to 2^+}{\lim} a + \dfrac{1-x}{2+x}$
$= a + \dfrac{-1}{4}$
Để hso ltuc tại $x = 2$ thì
$a -\dfrac{1}{4} = -1$
$<-> a = -\dfrac{3}{4}$.
Vậy $a = -\dfrac{3}{4}$.
