Giải thích các bước giải:

a.ĐKXĐ: $x\ge 0$

Để $A<0$

$\to \dfrac{2(\sqrt{x}-2)}{3(\sqrt{x}+1)}<0$

$\to \sqrt{x}-2<0$ vì $\sqrt{x}+1>0$

$\to \sqrt{x}<2$

$\to 0\le x<4$

b.Ta có:

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

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

$\to A=\dfrac{2(\sqrt{x}+1)-6}{3(\sqrt{x}+1)}$ 

$\to A=\dfrac23-\dfrac2{\sqrt{x}+1}$

Vì $x\in Z\to \sqrt{x}\in Z$ hoặc $\sqrt{x}\in I$

Nếu $\sqrt{x}\in I$

$\to \sqrt{x}+1\in I$

$\to \dfrac23-\dfrac2{\sqrt{x}+1}\in I\to \sqrt{x}\in I$ loại

$\to \sqrt{x}\in Z$

$\to 2(\sqrt{x}-2)\quad\vdots\quad 3(\sqrt{x}+1)$

$\to 2(\sqrt{x}-2)\quad\vdots\quad \sqrt{x}+1$

$\to 2(\sqrt{x}+1-3)\quad\vdots\quad \sqrt{x}+1$

$\to 2(\sqrt{x}+1)-6\quad\vdots\quad \sqrt{x}+1$

$\to 6\quad\vdots\quad \sqrt{x}+1$

$\to \sqrt{x}+1\in\{1, 2, 3, 6\}$

$\to \sqrt{x}\in\{0, 1, 2, 5\}$

$\to x\in\{0, 1, 4, 25\}$