1

Thay $x=9$ (thỏa mãn $x>0$) vào $A$, ta có:
$A = \dfrac{9+3}{2\sqrt{9}+4}$
$A = \dfrac{12}{2 \cdot 3 + 4}$
$A = \dfrac{12}{6+4}$
$A = \dfrac{12}{10}$
$A = \dfrac{6}{5}$

2

Điều kiện: $x>0, x \ne 4$.
Ta có:
$B = \left(\dfrac{\sqrt{x}}{x-4} + \dfrac{1}{\sqrt{x}-2}\right) : \dfrac{2}{\sqrt{x}-2}$
$B = \left(\dfrac{\sqrt{x}}{(\sqrt{x}-2)(\sqrt{x}+2)} + \dfrac{\sqrt{x}+2}{(\sqrt{x}-2)(\sqrt{x}+2)}\right) \cdot \dfrac{\sqrt{x}-2}{2}$
$B = \dfrac{\sqrt{x}+\sqrt{x}+2}{(\sqrt{x}-2)(\sqrt{x}+2)} \cdot \dfrac{\sqrt{x}-2}{2}$
$B = \dfrac{2\sqrt{x}+2}{(\sqrt{x}-2)(\sqrt{x}+2)} \cdot \dfrac{\sqrt{x}-2}{2}$
$B = \dfrac{2(\sqrt{x}+1)}{(\sqrt{x}-2)(\sqrt{x}+2)} \cdot \dfrac{\sqrt{x}-2}{2}$
$B = \dfrac{\sqrt{x}+1}{\sqrt{x}+2}$

3

3
$\dfrac{A}{B} = \dfrac{\dfrac{x+3}{2\sqrt{x}+4}}{\dfrac{\sqrt{x}+1}{\sqrt{x}+2}}$
$\dfrac{A}{B} = \dfrac{x+3}{2(\sqrt{x}+2)} \cdot \dfrac{\sqrt{x}+2}{\sqrt{x}+1}$
$\dfrac{A}{B} = \dfrac{x+3}{2(\sqrt{x}+1)}$
Theo đề bài, $\dfrac{A}{B} > \dfrac{3}{2}$
$\dfrac{x+3}{2(\sqrt{x}+1)} > \dfrac{3}{2}$
$\dfrac{x+3}{\sqrt{x}+1} > 3$
Vì $\sqrt{x}+1 > 0$ với mọi $x>0$, nên nhân chéo bpt
$x+3 > 3(\sqrt{x}+1)$
$x+3 > 3\sqrt{x}+3$
$x > 3\sqrt{x}$
$x - 3\sqrt{x} > 0$
$\sqrt{x}(\sqrt{x}-3) > 0$
Vì $\sqrt{x} > 0$ (do $x>0$)
`->`
$\sqrt{x}-3 > 0$
$\sqrt{x} > 3$
$x > 9$
Kết hợp với đk $x>0, x \ne 4$, ta được $x > 9$