$\begin{array}{l} A = \left( {\dfrac{{2\left( {x - 2\sqrt x  + 1} \right)}}{{x - 4}} - \dfrac{{2\sqrt x  - 1}}{{\sqrt x  + 2}}} \right):\dfrac{{\sqrt x }}{{\sqrt x  - 2}}\\ A = \left[ {\dfrac{{2\left( {x - 2\sqrt x  + 1} \right)}}{{\left( {\sqrt x  - 2} \right)\left( {\sqrt x  + 2} \right)}} - \dfrac{{2\sqrt x  - 1}}{{\sqrt x  + 2}}} \right].\dfrac{{\sqrt x  - 2}}{{\sqrt x }}\\ A = \dfrac{{2x - 4\sqrt x  + 2 - \left( {2\sqrt x  - 1} \right)\left( {\sqrt x  - 2} \right)}}{{\left( {\sqrt x  - 2} \right)\left( {\sqrt x  + 2} \right)}}.\dfrac{{\sqrt x  - 2}}{{\sqrt x }}\\ A = \dfrac{{2x - 4\sqrt x  + 2 - \left( {2x - 5\sqrt x  + 2} \right)}}{{\left( {\sqrt x  - 2} \right)\left( {\sqrt x  + 2} \right)}}.\dfrac{{\sqrt x  - 2}}{{\sqrt x }}\\ A = \dfrac{1}{{\sqrt x  + 2}} > \dfrac{1}{5}\\  \Rightarrow \sqrt x  + 2 < 5 \Rightarrow \sqrt x  < 3 \Rightarrow x < 9\\  \Rightarrow 0 < x < 9\\  \Rightarrow x \in \left\{ {1;2;3;5;6;7;8} \right\} \end{array}$  

$ĐKXĐ: x>0; x\ne4$

$A=[\dfrac{2(x-2\sqrt{x}+1)}{x-4}-\dfrac{2\sqrt{x}-1}{\sqrt{x}+2}]:\dfrac{\sqrt{x}}{\sqrt{x}-2}$

$A=[\dfrac{2x-4\sqrt{x}+2}{(\sqrt{x}+2)(\sqrt{x}-2)}-\dfrac{2\sqrt{x}-1}{\sqrt{x}+2}].\dfrac{\sqrt{x}-2}{\sqrt{x}}$

$A=\dfrac{(2x-4\sqrt{x}+2)-(2\sqrt{x}-1)(\sqrt{x}-2)}{(\sqrt{x}+2)(\sqrt{x}-2)}.\dfrac{\sqrt{x}-2}{\sqrt{x}}$

$A=\dfrac{2x-4\sqrt{x}+2-2x+5\sqrt{x}-2}{(\sqrt{x}+2)(\sqrt{x}-2)}.\dfrac{\sqrt{x}-2}{\sqrt{x}}$

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

$A=\dfrac{1}{\sqrt{x}+2}$

Vậy với $x>0; x\ne4$ thì $A=\dfrac{1}{\sqrt{x}+2}$

Để $A>\dfrac{1}{5}$

$\Leftrightarrow\dfrac{1}{\sqrt{x}+2}>\dfrac{1}{5}$

$\Leftrightarrow\dfrac{1}{\sqrt{x}+2}-\dfrac{1}{5}>0$

$\Leftrightarrow\dfrac{5-(\sqrt{x}+2)}{\sqrt{x}+2}>0$

$\Leftrightarrow\dfrac{5-\sqrt{x}-2}{\sqrt{x}+2}>0$

$\Leftrightarrow\dfrac{3-\sqrt{x}}{\sqrt{x}+2}>0$

$\Leftrightarrow3-\sqrt{x}>0($vì$: \sqrt{x}+2>2>0$ với $\forall x>0; x\ne4)$

$\Leftrightarrow-\sqrt{x}>-3$

$\Leftrightarrow\sqrt{x}<3$

$\Leftrightarrow x<9$

Kết hợp với $ĐKXĐ: x>0; x\ne4$ và $x$ nguyên

Vậy $x\in\{1; 2; 3; 5; 6; 7; 8\}$ thì $A>\dfrac{1}{5}$