Đáp án:
$\begin{array}{l}
3)x \ge 0;x \ne 1\\
\dfrac{A}{B} = m\\
\Rightarrow \dfrac{{5\sqrt x + 9}}{{\sqrt x + 1}} = m\\
\Rightarrow m\sqrt x + m = 5\sqrt x + 9\\
\Rightarrow \left( {m - 5} \right).\sqrt x = 9 - m\\
Do:x \ge 0;x \ne 1\\
\Rightarrow \sqrt x \ge 0;\sqrt x \ne 1\\
\Rightarrow \left\{ \begin{array}{l}
m - 5 \ne 0\\
\dfrac{{9 - m}}{{m - 5}} \ge 0\\
\dfrac{{9 - m}}{{m - 5}} \ne 1
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
m \ne 5\\
\dfrac{{m - 9}}{{m - 5}} \le 0\\
\dfrac{{9 - m - \left( {m - 5} \right)}}{{m - 5}} \ne 0
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
m \ne 5\\
5 < m \le 9\\
9 - m - m + 5 \ne 0
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
5 < m \le 9\\
m \ne 7
\end{array} \right.\\
Vậy\,5 < m \le 9;m \ne 7
\end{array}$
