Giải thích các bước giải:

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

$\to A=\dfrac{5+3}{5+1}=\dfrac43$

2.Ta có:

$B=\dfrac{\sqrt{x}}{\sqrt{x}-1}+\dfrac1{\sqrt{x}+2}-\dfrac{3\sqrt{x}}{x+\sqrt{x}-2}$

$\to B=\dfrac{\sqrt{x}}{\sqrt{x}-1}+\dfrac1{\sqrt{x}+2}-\dfrac{3\sqrt{x}}{(\sqrt{x}+2)(\sqrt{x}-1)}$

$\to B=\dfrac{\sqrt{x}(\sqrt{x}+2)+(\sqrt{x}-1)-3\sqrt{x}}{(\sqrt{x}-1)(\sqrt{x}+2)}$

$\to B=\dfrac{x-1}{(\sqrt{x}-1)(\sqrt{x}+2)}$

$\to B=\dfrac{(\sqrt{x}-1)(\sqrt{x}+1)}{(\sqrt{x}-1)(\sqrt{x}+2)}$

$\to B=\dfrac{\sqrt{x}+1}{\sqrt{x}+2}$

3.Ta có:

$P=AB$

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

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

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

$\to \sqrt{P}>1$

$\to \sqrt{P}-1>0$

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

$\to P-\sqrt{P}>0$

$\to P>\sqrt{P}$