Đáp án đúng: A Giải chi tiết:Điều kiện: \(x > 0,\,\,x \ne 4.\) \(P = \frac{B}{A} = \frac{{\sqrt x - 2}}{{\sqrt x }}:\frac{{2\sqrt x + 1}}{{\sqrt x }} = \frac{{\sqrt x - 2}}{{\sqrt x }}.\frac{{\sqrt x }}{{2\sqrt x + 1}} = \frac{{\sqrt x - 2}}{{2\sqrt x + 1}}.\) Do \(\left| P \right| = \left\{ {\begin{array}{*{20}{c}}P&{{\rm{khi}}\,\,{\rm{ P}} \ge {\rm{0}}}\\{ - P}&{{\rm{khi}}\,\,{\rm{ P < }}0}\end{array}} \right. \Rightarrow \left| P \right| > P \Leftrightarrow P < 0.\) \(P < 0 \Leftrightarrow \frac{{\sqrt x - 2}}{{2\sqrt x + 1}} < 0 \Leftrightarrow \sqrt x - 2 < 0\,\,\,\left( {do\,\,2\sqrt x + 1 > 0} \right) \Leftrightarrow \sqrt x < 2 \Leftrightarrow x < 4\) Vậy \(0 < x < 4\) thì \(\left| P \right| > P.\) Chọn A.