1) Ta có \(A = \frac{{\sqrt x  + 4}}{{\sqrt x  - 1}} = \frac{{\sqrt 9  + 4}}{{\sqrt 9  - 1}} = \frac{7}{2}\)

2) \(B = \frac{{3\sqrt x  + 1}}{{x + 2\sqrt x  - 3}} - \frac{2}{{\sqrt x  + 3}} = \frac{{3\sqrt x  + 1}}{{(\sqrt x  + 3)(\sqrt x  - 1)}} - \frac{2}{{\sqrt x  + 3}}\)

\( = \frac{{3\sqrt x  + 1 - 2\left( {\sqrt x  - 1} \right)}}{{\left( {\sqrt x  - 1} \right)\left( {\sqrt x  + 3} \right)}} = \frac{{\sqrt x  + 3}}{{\left( {\sqrt x  - 1} \right)\left( {\sqrt x  + 3} \right)}} = \frac{1}{{\sqrt x  - 1}}\)

3) \(\frac{A}{B} \ge \frac{x}{4} + 5\)

\(\begin{array}{l}
 \Leftrightarrow \frac{{\sqrt x  + 4}}{{\sqrt x  - 1}}:\frac{1}{{\sqrt x  - 1}} \ge \frac{x}{4} + 5 \Leftrightarrow \frac{{\sqrt x  + 4}}{{\sqrt x  - 1}}.\sqrt x  - 1 \ge \frac{x}{4} + 5\\
 \Leftrightarrow \sqrt x  + 4 \ge \frac{x}{4} + 5 \Leftrightarrow \frac{x}{4} - \sqrt x  + 1 \le 0\\
 \Leftrightarrow x - 4\sqrt x  + 4 \le 0 \Leftrightarrow {(\sqrt x  - 2)^2} \le 0
\end{array}\)

Mặt khác \({(\sqrt x  - 2)^2} \ge 0\) với mọi \(x \ge 0\)

Do đó \({(\sqrt x  - 2)^2} \le 0 \Leftrightarrow \sqrt x  - 2 = 0 \Leftrightarrow \sqrt x  = 2 \Leftrightarrow x = 4\)