$A = \sqrt{1 - \dfrac{1}{xy}}$

Ta có: $x, y \neq 0$

$\Rightarrow x^2 y^2 > 0$

$\Rightarrow x^3 + y^3 = 2x^2 y^2 > 0$

$\Rightarrow x^3 + y^3 \neq 0$

Xét: $1 - \left(\dfrac{x^3 - y^3}{x^3 + y^3}\right)^2$

$= \dfrac{(x^3 + y^3)^2 - (x^3 - y^3)^2}{(x^3 + y^3)^2}$

$= \dfrac{[(x^3 + y^3) - (x^3 - y^3)][(x^3 + y^3) + (x^3 - y^3)]}{(x^3 + y^3)^2}$

$= \dfrac{2y^3 \cdot 2x^3}{(x^3 + y^3)^2}$

$= \dfrac{4x^3 y^3}{(x^3 + y^3)^2}$

$= \dfrac{4x^3 y^3}{(2x^2 y^2)^2}$

$= \dfrac{4x^3 y^3}{4x^4 y^4}$

$= \dfrac{1}{xy}$

$\Rightarrow 1 - \dfrac{1}{xy} = \left(\dfrac{x^3 - y^3}{x^3 + y^3}\right)^2$

Từ đó $A$ trở thành:

$A = \sqrt{\left(\dfrac{x^3 - y^3}{x^3 + y^3}\right)^2}$

$A = \left|\dfrac{x^3 - y^3}{x^3 + y^3}\right|$

Vì $\begin{cases} x \in \mathbb{Q} \\ y \in \mathbb{Q} \end{cases}$

$\Rightarrow \begin{cases} x^3 - y^3 \in \mathbb{Q} \\ x^3 + y^3 \in \mathbb{Q} \end{cases}$

$\Rightarrow \dfrac{x^3 - y^3}{x^3 + y^3} \in \mathbb{Q}$

$\Rightarrow \left|\dfrac{x^3 - y^3}{x^3 + y^3}\right| \in \mathbb{Q}$

$\Rightarrow A \in \mathbb{Q}$ (đpcm)

$\color{#FF0000}{♡^♡}\color{#FF7F00}{𝖘}\color{#FFD700}{𝖒}\color{#00FF00}{𝖕}\color{#00BFFF}{9}\color{#0000FF}{9}\color{#9400D3}{9}\color{#FF0000}{♡^♡}$