Đáp án:
$\begin{array}{l}
Dkxd;x \ge 0;x \ne 4;x \ne 9\\
A = \frac{{\sqrt x }}{{3 - \sqrt x }};B = \frac{{2\sqrt x + 1}}{{3 - \sqrt x }}\\
A < B\\
\Rightarrow \frac{{\sqrt x }}{{3 - \sqrt x }} < \frac{{2\sqrt x + 1}}{{3 - \sqrt x }}\\
\Rightarrow \frac{{\sqrt x - 2\sqrt x + 1}}{{3 - \sqrt x }} < 0\\
\Rightarrow \frac{{1 - \sqrt x }}{{3 - \sqrt x }} < 0\\
\Rightarrow \frac{{\sqrt x - 1}}{{\sqrt x - 3}} < 0\\
\Rightarrow 1 < \sqrt x < 3\\
\Rightarrow 1 < x < 9\\
Vậy\,1 < x < 9;x \ne 4
\end{array}$