Đáp án:
\(a = - \frac{1}{6}\)
Giải thích các bước giải:
Để hàm số liên tục tại x=0
\(\begin{array}{l}
\Leftrightarrow \mathop {\lim }\limits_{x \to 0} f\left( x \right) = f\left( 0 \right)\\
Có:\mathop {\lim }\limits_{x \to 0} f\left( x \right) = \mathop {\lim }\limits_{x \to 0} \frac{{4x + 1 - 1}}{{x\left( {ax + 2a + 1} \right)\left( {\sqrt {4x + 1} + 1} \right)}}\\
= \mathop {\lim }\limits_{x \to 0} \frac{{4x}}{{x\left( {ax + 2a + 1} \right)\left( {\sqrt {4x + 1} + 1} \right)}}\\
= \mathop {\lim }\limits_{x \to 0} \frac{4}{{\left( {ax + 2a + 1} \right)\left( {\sqrt {4x + 1} + 1} \right)}}\\
= \frac{4}{{\left( {a.0 + 2a + 1} \right)\left( {\sqrt {4.0 + 1} + 1} \right)}}\\
= \frac{4}{{2\left( {2a + 1} \right)}}\\
f\left( 0 \right) = 3\\
\to \frac{4}{{2\left( {2a + 1} \right)}} = 3\left( {a \ne - \frac{1}{2}} \right)\\
\to 4 = 6\left( {2a + 1} \right)\\
\to 12a + 6 = 4\\
\to 12a = - 2\\
\to a = - \frac{1}{6}\left( {TM} \right)
\end{array}\)
