Đáp án:
\(a = - \frac{{11}}{6}\)
Giải thích các bước giải:
Do hàm số liên tục tại x=4
\(\begin{array}{l}
\to \mathop {\lim }\limits_{x \to 4} f\left( x \right) = f\left( 4 \right)\\
\mathop {\lim }\limits_{x \to 4} f\left( x \right) = \mathop {\lim }\limits_{x \to 4} \frac{{\sqrt {2x + 1} - \sqrt {x + 5} }}{{x - 4}}\\
= \mathop {\lim }\limits_{x \to 4} \frac{{2x + 1 - x - 5}}{{\left( {x - 4} \right)\left( {\sqrt {2x + 1} + \sqrt {x + 5} } \right)}}\\
= \mathop {\lim }\limits_{x \to 4} \frac{{x - 4}}{{\left( {x - 4} \right)\left( {\sqrt {2x + 1} + \sqrt {x + 5} } \right)}}\\
= \mathop {\lim }\limits_{x \to 4} \frac{1}{{\sqrt {2x + 1} + \sqrt {x + 5} }}\\
= \frac{1}{{3 + 3}} = \frac{1}{6}\\
f\left( 4 \right) = a + 2\\
\to a + 2 = \frac{1}{6}\\
\to a = - \frac{{11}}{6}
\end{array}\)
