Đáp án:
\(m = - \dfrac{9}{4}\)
Giải thích các bước giải:
\(\begin{array}{l}
Xét:\\
\mathop {\lim }\limits_{x \to 1} f\left( x \right) = \mathop {\lim }\limits_{x \to 1} \dfrac{{\sqrt x - 3 + \sqrt {x + 3} }}{{x - 1}}\\
= \mathop {\lim }\limits_{x \to 1} \dfrac{{x - 6\sqrt x + 9 - x - 3}}{{\left( {x - 1} \right)\left( {\sqrt x - 3 - \sqrt {x + 3} } \right)}}\\
= \mathop {\lim }\limits_{x \to 1} \dfrac{{ - 6\sqrt x + 6}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)\left( {\sqrt x - 3 - \sqrt {x + 3} } \right)}}\\
= \mathop {\lim }\limits_{x \to 1} \dfrac{{ - 6\left( {\sqrt x - 1} \right)}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)\left( {\sqrt x - 3 - \sqrt {x + 3} } \right)}}\\
= \mathop {\lim }\limits_{x \to 1} \dfrac{{ - 6}}{{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 3 - \sqrt {x + 3} } \right)}}\\
= \dfrac{{ - 6}}{{\left( {\sqrt 1 + 1} \right)\left( {\sqrt 1 - 3 - \sqrt {1 + 3} } \right)}} = \dfrac{{ - 6}}{{ - 8}} = \dfrac{3}{4}\\
\mathop {\lim }\limits_{x \to 1} f\left( x \right) = \mathop {\lim }\limits_{x \to 1} \left( {3 + m} \right) = 3 + m
\end{array}\)
Để hàm số liên tục tại x=1
\(\begin{array}{l}
\to 3 + m = \dfrac{3}{4}\\
\to m = - \dfrac{9}{4}
\end{array}\)
