Giải thích các bước giải:

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

$\text{ĐKXĐ: $x > 0$ và $x \neq 1$}$

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

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

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

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

⇔ $A=\dfrac{2\sqrt{x}(x-1)}{2\sqrt{x}.(\sqrt{x}-1)^2}$

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

⇔ $A=\dfrac{\sqrt{x}+1}{\sqrt{x}-1}$

2/ $\text{Để $A < 0$ thì $\dfrac{\sqrt{x}+1}{\sqrt{x}-1} < 0$}$ 

$\text{Mà $\sqrt{x} \geq 0$ nên $\sqrt{x}+1 > 0$}$

$\text{⇒ $\sqrt{x}-1 < 0$}$

$\text{⇒ $\sqrt{x} < 1$}$

$\text{⇒ $x < 1$}$

$\text{Vậy để $A<0$ thì $0 < x < 1$}$

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

$\text{Để A nguyên thì $\dfrac{2}{\sqrt{x}-1}$ nguyên}$

$\text{⇔ $2 \vdots (\sqrt{x}-1)$}$

$\text{⇔ $(\sqrt{x}-1) ∈ Ư_{(2)}=${1; -1; 2; -2}}$

$\text{*) $\sqrt{x}-1=1$ ⇔ $\sqrt{x}=2$ ⇔ $x=4$ (nhận)}$

$\text{*) $\sqrt{x}-1=-1$ ⇔ $\sqrt{x}=0$ ⇔ $x=0$ (KTM ĐKXĐ)}$

$\text{*) $\sqrt{x}-1=2$ ⇔ $\sqrt{x}=3$ ⇔ $x=9$ (nhận)}$

$\text{*) $\sqrt{x}-1=-2$ ⇔ $\sqrt{x}=-1$ (loại)}$

$\text{Vậy}$ `x={4; 9}`

Chúc bạn học tốt !!!

$\begin{array}{l}A = \left(\dfrac{x\sqrt x -1}{x - \sqrt x}- \dfrac{x\sqrt x + 1}{x + \sqrt x}\right):\dfrac{2(x - 2\sqrt x + 1)}{x - 1}\\ ĐKXĐ: \, x >0;\, x \ne 1\\ A = \left[\dfrac{(\sqrt x - 1)(x + \sqrt x + 1)}{\sqrt x(\sqrt x - 1)}- \dfrac{(\sqrt x + 1)(x - \sqrt x + 1)}{\sqrt x(\sqrt x + 1)}\right]:\dfrac{2(\sqrt x - 1)^2}{(\sqrt x - 1)(\sqrt x + 1)}\\ =\left[\dfrac{x + \sqrt x + 1- (x - \sqrt x + 1)}{\sqrt x}\right]:\dfrac{2(\sqrt x - 1)}{\sqrt x + 1}\\ = \dfrac{2\sqrt x}{\sqrt x}\cdot \dfrac{\sqrt x+1}{2(\sqrt x - 1)}\\ = \dfrac{\sqrt x + 1}{\sqrt x - 1}\\ Ta \,\,có:\\ \sqrt x > 0\\ \Leftrightarrow \sqrt x + 1 > 0\\ \Rightarrow A < 0\\ \Leftrightarrow \dfrac{\sqrt x + 1}{\sqrt x - 1} < 0\\ \Leftrightarrow \sqrt x - 1 < 0\\ \Leftrightarrow \sqrt x < 1\\ \Leftrightarrow x < 1\\ \text{Kết hợp điều kiện xác định, ta được:}\\ 0 < x < 1\\ Ta\,\, có:\\ A = \dfrac{\sqrt x + 1}{\sqrt x - 1} = \dfrac{\sqrt x - 1 + 2}{\sqrt x - 1} = 1 + \dfrac{2}{\sqrt x - 1}\\ A \in \Bbb Z \Leftrightarrow \dfrac{2}{\sqrt x - 1} \Leftrightarrow \sqrt x - 1 \in Ư(2) = \left\{-2;-1;1;2\right\}\\Do\,\,\sqrt x > 0\\nên\,\,\sqrt x - 1 > -1\\\Rightarrow \sqrt x - 1 = \left\{1;2\right\}\\ \text{Ta có bảng giá trị:}\\\begin{array}{|l|r|}
\hline
\sqrt x - 1 &1\quad &2\quad \\
\hline
\quad \sqrt x &2\quad&3\quad \\
\hline
\quad x&4\quad &9\quad\\
\hline
\end{array}\end{array}$