Đáp án + Giải thích các bước giải:

$\textbf{1}\big/$

$\begin{array}{l} \textbf{P} &= \dfrac{\sqrt{x}}{\sqrt{x} - 2} : \bigg(\dfrac{x - 2}{x - 4} - \dfrac{1}{\sqrt{x} + 2}\bigg) (x > 0, x \ne 1; 4)
\\ &= \dfrac{\sqrt{x}}{\sqrt{x} - 2} : \bigg(\dfrac{x - 2}{(\sqrt{x} + 2)(\sqrt{x} - 2)} - \dfrac{1}{\sqrt{x} + 2}\bigg) 
\\ &= \dfrac{\sqrt{x}}{\sqrt{x} - 2} : \bigg(\dfrac{x - 2}{(\sqrt{x} + 2)(\sqrt{x} - 2)} - \dfrac{\sqrt{x} - 2}{(\sqrt{x} + 2)(\sqrt{x} - 2)}\bigg)
\\ &= \dfrac{\sqrt{x}}{\sqrt{x} - 2} : \dfrac{x - \sqrt{x}}{(\sqrt{x} + 2)(\sqrt{x} - 2)}
\\ &= \dfrac{\sqrt{x}}{\sqrt{x} - 2} \cdot \dfrac{(\sqrt{x} + 2)(\sqrt{x} - 2)}{\sqrt{x}(\sqrt{x} - 1)}
\\ &= \dfrac{\sqrt{x} + 2}{\sqrt{x} - 1} \end{array}$

$\textbf{2}\big/$

$\begin{array}{l} \textbf{P} &= \dfrac{\sqrt{x} + 2}{\sqrt{x} - 1}
\\&= \dfrac{\sqrt{x} - 1 + 3}{\sqrt{x} - 1}
\\&= 1 + \dfrac{3}{\sqrt{x} - 1}\end{array}$

$\textbf{P}  = 1 + \dfrac{3}{\sqrt{x} - 1} > 1$

$\dfrac{3}{\sqrt{x} -1 } > 0$

$\sqrt{x} -1  > 0$

$\sqrt{x} > 1$

$x > 1$

Vậy với $x >1$ và $x \ne 4$ thì $\textbf{P} > 1$

$\textbf{3}\big/$

Thay $x = \dfrac{1}{4}$ vào $\textbf{P} = 1 + \dfrac{3}{\sqrt{x} - 1}$, ta được:

$\textbf{P} = 1 + \dfrac{3}{\sqrt{\frac{1}{4}} - 1} = 1 + \dfrac{3}{-\frac{1}{2}} = -5$

Vậy khi $x = \dfrac{1}{4}$ thì $\textbf{P} = -5$