Giải thích các bước giải:

1.Khi $x=25\to \sqrt{x}=5$

$\to A=\dfrac{5-3}5=\dfrac25$ 

2.Ta có:

$B=\dfrac{\sqrt{x}}{\sqrt{x}+2}+\dfrac2{2-\sqrt{x}}+\dfrac{2\sqrt{x}+4}{x-4}$

$\to B=\dfrac{\sqrt{x}}{\sqrt{x}+2}-\dfrac2{\sqrt{x}-2}+\dfrac{2(\sqrt{x}+2)}{(\sqrt{x}-2)(\sqrt{x}+2)}$

$\to B=\dfrac{\sqrt{x}}{\sqrt{x}+2}-\dfrac2{\sqrt{x}-2}+\dfrac2{\sqrt{x}-2}$

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

3.Ta có:

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

Để $\sqrt{P}<\dfrac23$

$\to 0\le P<\dfrac49$

$\to 0\le\dfrac{\sqrt{x}-3}{\sqrt{x}+2}<\dfrac49$

$\to 3\le \sqrt{x}<7$

$\to 9\le x<49$