Giải thích các bước giải:

1.Khi $x=9\to \sqrt{x}=3$

$\to M=\dfrac{9+3}{3-2}=12$

2.Ta có:

$N=\dfrac{\sqrt{x}-1}{\sqrt{x}+2}+\dfrac{5\sqrt{x}-2}{x-4}$

$\to N=\dfrac{(\sqrt{x}-1)(\sqrt{x}-2)+5\sqrt{x}-2}{(\sqrt{x}-2)(\sqrt{x}+2)}$ 

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

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

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

3.Ta có:

$\dfrac{M}{N}=\dfrac{x+3}{\sqrt{x}-2}:\dfrac{\sqrt{x}}{\sqrt{x}-2}$

$\to \dfrac{M}N=\dfrac{x+3}{\sqrt{x}}=\sqrt{x}+\dfrac3{\sqrt{x}}\ge 2\sqrt{\sqrt{x}\cdot\dfrac3{\sqrt{x}}}=2\sqrt3$

Dấu = xảy ra khi $\sqrt{x}=\dfrac3{\sqrt{x}}\to x=3$