Giải thích các bước giải:

1.Khi $x=16\to \sqrt{x}=4$

$\to A=\dfrac{16-3}{4+3}=\dfrac{13}7$

2.Ta có:

$B=(\dfrac{x+3\sqrt{x}-2}{x-9}-\dfrac1{\sqrt{x}+3})\cdot\dfrac{\sqrt{x}-3}{\sqrt{x}+1}$

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

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

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

$\to B=\dfrac{\sqrt{x}+1}{\sqrt{x}+3}$

3.Ta có:

$P=\dfrac{B}A$

$\to P=\dfrac{\sqrt{x}+1}{\sqrt{x}+3}:\dfrac{x-3}{\sqrt{x}+3}$

$\to P=\dfrac{\sqrt{x}+1}{x-3}$

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

Nếu $\sqrt{x}\in I$

$\to \dfrac{\sqrt{x}+1}{x-3}\in I$

$\to \sqrt{x}\in I$ loại

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

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

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

$\to x-1\quad\vdots\quad x-3$

$\to x-3+2\quad\vdots\quad x-3$

$\to 2\quad\vdots\quad x-3$

$\to x-3\in U(2)$

$\to x-3\in \{1, 2, -1, -2\}$

$\to x\in\{4, 5, 2, 1\}$