`1)`

$A = \dfrac{\sqrt{x}+5}{\sqrt{x}-2}$ với $x \ge 0; x \ne 4$

Thay $x = \dfrac{1}{4}$ (TMĐK) vào `A` ta có: 

$A = \dfrac{\sqrt{\dfrac{1}{4}}+5}{\sqrt{\dfrac{1}{4}}-2} = \dfrac{\dfrac{1}{2}+5}{\dfrac{1}{2}-2} = \dfrac{\dfrac{11}{2}}{-\dfrac{3}{2}} = -\dfrac{11}{3}$

Vậy $A = -\dfrac{11}{3}$ khi $x = \dfrac{1}{4}$

` 2)` 

$B = \dfrac{\sqrt{x}}{\sqrt{x}+2} - \dfrac{3}{2-\sqrt{x}} + \dfrac{3\sqrt{x}-2}{x-4}$ với $x \ge 0; x \ne 4$

$B = \dfrac{\sqrt{x}}{\sqrt{x}+2} + \dfrac{3}{\sqrt{x}-2} + \dfrac{3\sqrt{x}-2}{(\sqrt{x}-2)(\sqrt{x}+2)}$

$B = \dfrac{\sqrt{x}(\sqrt{x}-2) + 3(\sqrt{x}+2) + 3\sqrt{x}-2}{(\sqrt{x}-2)(\sqrt{x}+2)}$

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

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

$B = \dfrac{(\sqrt{x}+2)^2}{(\sqrt{x}-2)(\sqrt{x}+2)}$

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

Vậy $B = \dfrac{\sqrt{x}+2}{\sqrt{x}-2}$ với $x \ge 0; x \ne 4$

`3)`

$A = \dfrac{\sqrt{x}+5}{\sqrt{x}-2}$; $B = \dfrac{\sqrt{x}+2}{\sqrt{x}-2}$ với $x \ge 0; x \ne 4$

Ta có: $P = A : B$ 

$P = \dfrac{\sqrt{x}+5}{\sqrt{x}-2} : \dfrac{\sqrt{x}+2}{\sqrt{x}-2}$

$P = \dfrac{\sqrt{x}+5}{\sqrt{x}-2} \cdot \dfrac{\sqrt{x}-2}{\sqrt{x}+2}$

$P = \dfrac{\sqrt{x}+5}{\sqrt{x}+2}$

$P = \dfrac{\sqrt{x}+2+3}{\sqrt{x}+2}$

$P = 1 + \dfrac{3}{\sqrt{x}+2}$

Vì $x \ge 0 \Rightarrow \sqrt{x} \ge 0 \Rightarrow \sqrt{x}+2 \ge 2 > 0$

$\Rightarrow \dfrac{3}{\sqrt{x}+2} > 0$

$\Rightarrow P > 1$

$\Rightarrow \sqrt{P} > 1$

$\Rightarrow P - \sqrt{P} = \sqrt{P}(\sqrt{P} - 1) > 0$

$\Rightarrow P > \sqrt{P}$

Vậy $P > \sqrt{P}$

$\color{#FF0000}{♡^♡}\color{#FF7F00}{𝖘}\color{#FFD700}{𝖒}\color{#00FF00}{𝖕}\color{#00BFFF}{9}\color{#0000FF}{9}\color{#9400D3}{9}\color{#FF0000}{♡^♡}$