Đáp án+  Giải thích các bước giải:

Bài $\textbf{II}.$

$\textbf{1}\bigg)$

Thay $x = \dfrac{1}{4}$ vào $\textbf{A}$, ta có:

$\textbf{A} = \dfrac{2}{\sqrt{\frac{1}{4}} - 2} = \dfrac{2}{\frac{1}{2} - 2} = \dfrac{2}{-\frac{3}{2}} = -\dfrac{4}{3}$

Vậy với $x = \dfrac{1}{4}$ thì $\textbf{A} = -\dfrac{4}{3}$

$\textbf{2}\bigg)$

$\textbf{B} = \dfrac{3}{\sqrt{x} - 2} + \dfrac{\sqrt{x}+ 1}{\sqrt{x} + 2} - \dfrac{2\sqrt{x}}{4 - x} (x \ge 0, x \ne 4)$

$\phantom{\textbf{B}} = \dfrac{3}{\sqrt{x} - 2} +\dfrac{\sqrt{x} + 1}{\sqrt{x} + 2} + \dfrac{2\sqrt{x}}{x - 4}$

$\phantom{\textbf{B}} = \dfrac{3}{\sqrt{x} - 2} +\dfrac{\sqrt{x} + 1}{\sqrt{x} + 2} + \dfrac{2\sqrt{x}}{\left(\sqrt{x} + 2\right)\left(\sqrt{x} - 2\right)}$

$\phantom{\textbf{B}} = \dfrac{3\left(\sqrt{x}+  2\right)}{\left(\sqrt{x} + 2\right)\left(\sqrt{x} - 2\right)} +\dfrac{\left(\sqrt{x} + 1\right)\left(\sqrt{x}-  2\right)}{\left(\sqrt{x} + 2\right)\left(\sqrt{x} - 2\right)} + \dfrac{2\sqrt{x}}{\left(\sqrt{x} + 2\right)\left(\sqrt{x} - 2\right)}$

$\phantom{\textbf{B}} = \dfrac{3\left(\sqrt{x}+  2\right) + \left(\sqrt{x} + 1\right)\left(\sqrt{x}-  2\right) + 2\sqrt{x}}{\left(\sqrt{x} + 2\right)\left(\sqrt{x} - 2\right)}$

$\phantom{\textbf{B}} = \dfrac{3\sqrt{x}+  6 + x + \sqrt{x} - 2\sqrt{x} - 2 + 2\sqrt{x}}{\left(\sqrt{x} + 2\right)\left(\sqrt{x} - 2\right)}$

$\phantom{\textbf{B}} = \dfrac{x + 4\sqrt{x} + 4}{\left(\sqrt{x} + 2\right)\left(\sqrt{x} - 2\right)}$

$\phantom{\textbf{B}} = \dfrac{\left(\sqrt{x} + 2\right)^2}{\left(\sqrt{x} + 2\right)\left(\sqrt{x} - 2\right)}$

$\phantom{\textbf{B}} = \dfrac{\sqrt{x} + 2}{\sqrt{x} - 2}$

$\textbf{3}\bigg)$

$\textbf{P} = \dfrac{\textbf{A}}{\textbf{B}}$

$\phantom{\textbf{P}} =\dfrac{2}{\sqrt{x} - 2} : \dfrac{\sqrt{x} + 2}{\sqrt{x} -2}$

$\phantom{\textbf{P}} =\dfrac{2}{\sqrt{x} - 2} \cdot \dfrac{\sqrt{x} -2}{\sqrt{x} + 2}$

$\phantom{\textbf{P}} =\dfrac{2}{\sqrt{x} + 2}$

Ta có: $\textbf{P} =\dfrac{2}{\sqrt{x} + 2} \ge \dfrac{2}{x + 2}$

Suy ra $\sqrt{x} + 2  \le x + 2$

$\sqrt{x} \le x$

$\sqrt{x} - x\le 0$

$\sqrt{x}\left(1 - \sqrt{x}\right) \le 0$

$\begin {cases} \sqrt{x} \le 0 \\ 1 - \sqrt{x} \ge 0\end {cases}$ hoặc $\begin {cases} \sqrt{x} \ge 0 \\ 1 - \sqrt{x} \le 0\end {cases}$

$\begin {cases} x = 0 \\ \sqrt{x} \le 1 \end {cases}$ hoặc $\begin {cases} x \ge 0 \\ \sqrt{x} \ge 1 \end {cases}$

$\begin {cases} x = 0 \\ 0 \le x \le 1 \end {cases}$ hoặc $\begin {cases} x \ge 0 \\ x \ge 1 \end {cases}$

$x = 0$ hoặc $x \ge 1$

Vậy $x = 0$ hoặc $x \ge 1$