$\lim\limits_{x\to 2^+}f(x)$
$=\lim\limits_{x\to 2^+}\dfrac{ \sqrt{7x-10}-2}{x-2}$
$=\lim\limits_{x\to 2^+}\dfrac{7x-10-4}{(x-2)(\sqrt{7x-10}+2}$
$=\lim\limits_{x\to 2^+}\dfrac{7}{\sqrt{7x-10}+2}$
$=\dfrac{7}{\sqrt{7.2-10}+2}$
$=\dfrac{7}{4}$
$\Rightarrow \lim\limits_{x\to 2^-}f(x)=\dfrac{7}{4}$
$\Leftrightarrow \lim\limits_{x\to 2^-}(mx+3)=\dfrac{7}{4}$
$\Leftrightarrow 2m+3=\dfrac{7}{4}$
$\Leftrightarrow m=\dfrac{-5}{8}$
Khi đó $f(2)=\dfrac{-5}{8}.2+3=\dfrac{7}{4}=\lim\limits_{x\to 2}f(x)$ nên $f(x)$ liên tục tại $x=2$
