Đáp án:
$m=-\dfrac{5}{3}$
Giải thích các bước giải:
Đề hàm số liên tục tại $x=2$
$\displaystyle\lim_{x \to 2^-} f(x)=\displaystyle\lim_{x \to 2^+} f(x)\\ \Leftrightarrow \displaystyle\lim_{x \to 2^-} \dfrac{\sqrt{x+7}-\sqrt{3x+3}}{x-2}=\displaystyle\lim_{x \to 2^+} \dfrac{2mx-3m+2}{x-3}\\ \Leftrightarrow \displaystyle\lim_{x \to 2^-} \dfrac{\left(\sqrt{x+7}-\sqrt{3x+3}\right)\left(\sqrt{x+7}+\sqrt{3x+3}\right)}{(x-2)\left(\sqrt{x+7}+\sqrt{3x+3}\right)}=\displaystyle\lim_{x \to 2^+} \dfrac{2.2m-3m+2}{2-3}\\ \Leftrightarrow \displaystyle\lim_{x \to 2^-} \dfrac{x+7-3x-3}{(x-2)\left(\sqrt{x+7}+\sqrt{3x+3}\right)}=-m-2\\ \Leftrightarrow \displaystyle\lim_{x \to 2^-} \dfrac{-2x+4}{(x-2)\left(\sqrt{x+7}+\sqrt{3x+3}\right)}=-m-2\\ \Leftrightarrow \displaystyle\lim_{x \to 2^-} \dfrac{-2}{\sqrt{x+7}+\sqrt{3x+3}}=-m-2\\ \Leftrightarrow \dfrac{-2}{\sqrt{2+7}+\sqrt{3.2+3}}=-m-2\\ \Leftrightarrow m=-\dfrac{5}{3}$
