Giải thích các bước giải:

$1) A = \left ( \dfrac{\sqrt{x} - 1}{x - 4} - \dfrac{\sqrt{x} + 1}{x + 4\sqrt{x} + 4}  \right ) : \dfrac{4\sqrt{x}}{\left ( 4 - x \right )^{2}}$ (ĐK $x > 0, x \neq 4$)

$= \left [ \dfrac{\sqrt{x} - 1}{\left ( \sqrt{x} + 2 \right )\left ( \sqrt{x} - 2 \right )} - \dfrac{\sqrt{x} + 1}{\left ( \sqrt{x} + 2 \right )^{2}} \right ] : \dfrac{4\sqrt{x}}{\left ( 4 - x \right )^{2}}$

$= \left [ \dfrac{\left ( \sqrt{x} - 1 \right )\left ( \sqrt{x} + 2 \right )}{\left ( \sqrt{x} + 2 \right )^{2}\left ( \sqrt{x} - 2 \right )} - \dfrac{\left ( \sqrt{x} + 1 \right )\left ( \sqrt{x} - 2 \right )}{\left ( \sqrt{x} + 2 \right )^{2}\left ( \sqrt{x} - 2 \right )} \right ] : \dfrac{4\sqrt{x}}{\left ( 4 - x \right )^{2}}$

$= \dfrac{\left ( \sqrt{x} - 1 \right )\left ( \sqrt{x} + 2 \right ) - \left ( \sqrt{x} + 1 \right )\left ( \sqrt{x} - 2 \right )}{\left ( \sqrt{x} + 2 \right )^{2}\left ( \sqrt{x} - 2 \right )} : \dfrac{4\sqrt{x}}{\left ( 4 - x \right )^{2}}$

$= \dfrac{x + \sqrt{x} - 2 - x + \sqrt{x} + 2}{\left ( \sqrt{x} + 2 \right )^{2}\left ( \sqrt{x} - 2 \right )} : \dfrac{4\sqrt{x}}{\left ( 4 - x \right )^{2}}$

$= \dfrac{2\sqrt{x}}{\left ( \sqrt{x} + 2 \right )\left ( \sqrt{x} + 2 \right )\left ( \sqrt{x} - 2 \right )} : \dfrac{4\sqrt{x}}{\left ( 4 - x \right )^{2}}$

$= \dfrac{2\sqrt{x}}{\left ( \sqrt{x} + 2 \right )\left ( x - 4 \right )}.\dfrac{\left ( x - 4 \right )^{2}}{4\sqrt{x}}$

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

$= \dfrac{\left ( \sqrt{x} - 2 \right )\left ( \sqrt{x} + 2 \right )}{2\left ( \sqrt{x} + 2 \right )}$

$= \dfrac{\sqrt{x} - 2}{2}$

$2)$ Ta có $x = 4 + 2\sqrt{3} > 0$ và $x = 4 + 2\sqrt{3} \neq 4$ nên thay vào $A$ ta có:

$\dfrac{\sqrt{x} - 2}{2} = \dfrac{\sqrt{4 + 2\sqrt{3}} - 2}{2} = \dfrac{\sqrt{3 + 2\sqrt{3} + 1} - 2}{2} = \dfrac{\left ( \sqrt{3} + 1 \right )^{2} - 2}{2} = \dfrac{\sqrt{3} + 1 - 2}{2} = \dfrac{\sqrt{3} - 1}{2}$

$3)$ Với $x > 0, x \neq 4$ thì $A \geq \dfrac{1}{4}$ khi và chỉ khi:

$\dfrac{\sqrt{x} - 2}{2} \geq \dfrac{1}{4}$

$\Leftrightarrow \dfrac{\sqrt{x} - 2}{2} - \dfrac{1}{4} \geq 0$

$\Leftrightarrow \dfrac{2\left ( \sqrt{x} - 2 \right ) - 1}{4} \geq 0$

$\Leftrightarrow 2\sqrt{x} - 4 - 1 \geq 0$

$\Leftrightarrow 2\sqrt{x} - 5 \geq 0$

$\Leftrightarrow 2\sqrt{x} \geq 5$

$\Leftrightarrow \sqrt{x} \geq \dfrac{5}{2}$

$\Leftrightarrow x \geq \dfrac{25}{4}$

Vậy $x \geq \dfrac{25}{4}$ thì $A \geq \dfrac{1}{4}$