Giải thích các bước giải:

Ta có:

$A=\dfrac{2x-8\sqrt{x}+3}{x+\sqrt{x}-6}+\dfrac{\sqrt{x-2\sqrt{x}+1}}{\sqrt{x}-2}-\dfrac{2\sqrt{x}}{\sqrt{x}+3}$

$\to A=\dfrac{2x-8\sqrt{x}+3}{(\sqrt{x}+3)(\sqrt{x}-2)}+\dfrac{\sqrt{(\sqrt{x}-1)^2}}{\sqrt{x}-2}-\dfrac{2\sqrt{x}}{\sqrt{x}+3}$

$\to A=\dfrac{2x-8\sqrt{x}+3}{(\sqrt{x}+3)(\sqrt{x}-2)}+\dfrac{ (\sqrt{x}-1}{\sqrt{x}-2}-\dfrac{2\sqrt{x}}{\sqrt{x}+3}\text{ vì x>4}$

$\to A=\dfrac{2x-8\sqrt{x}+3+(\sqrt{x}-1)(\sqrt{x}+3)-2\sqrt{x}(\sqrt{x}-2)}{(\sqrt{x}+3)(\sqrt{x}-2)}$

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

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

$\to A=\dfrac{\sqrt{x}}{\sqrt{x}+3}$

Để $A\ge\dfrac{x-\sqrt{x}}{12}$

$\to \dfrac{\sqrt{x}}{\sqrt{x}+3}\ge\dfrac{x-\sqrt{x}}{12}$

$\to 12\sqrt{x}\ge (x-\sqrt{x})(\sqrt{x}+3)$

$\to 12\sqrt{x}\ge \sqrt{x}(\sqrt{x}-1)(\sqrt{x}+3)$

$\to \sqrt{x}(\sqrt{x}-1)(\sqrt{x}+3)-12\sqrt{x}\le 0$

$\to \sqrt{x}((\sqrt{x}-1)(\sqrt{x}+3)-12)\le 0$

$\to \sqrt{x}(x+2\sqrt{x}-15)\le 0$

$\to \sqrt{x}(\sqrt{x}+5)(\sqrt{x}-3)\le 0$ 

$\to \sqrt{x}(\sqrt{x}-3)\le 0$

$\to 0\le \sqrt{x}\le 3$

$\to 0\le x\le 9$

Mà $x>4$

$\to 4<x\le 9, x\in Z$

$\to x\in\{5, 6, 7, 8,9\}$