EM THAM KHẢO :

ĐKXĐ : $x > 0$

$P = A + 2B$

$= \dfrac{x - 27}{x + 3\sqrt{x}} + 2\left(\dfrac{\sqrt{x}}{\sqrt{x} + 3}\right)$

$= \dfrac{x - 27}{\sqrt{x}(\sqrt{x} + 3)} + \dfrac{2x}{\sqrt{x}(\sqrt{x} + 3)}$

$= \dfrac{x - 27 + 2x}{\sqrt{x}(\sqrt{x} + 3)}$

$= \dfrac{3x - 27}{\sqrt{x}(\sqrt{x} + 3)}$

$= \dfrac{3(x - 9)}{\sqrt{x}(\sqrt{x} + 3)}$

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

$= \dfrac{3(\sqrt{x} - 3)}{\sqrt{x}}$

$= \dfrac{3\sqrt{x} - 9}{\sqrt{x}}$

$= 3 - \dfrac{9}{\sqrt{x}}$

Để $P \in \mathbb{Z}$ với $x \in \mathbb{Z}$ thì $\dfrac{9}{\sqrt{x}} \in \mathbb{Z}$

$\Rightarrow \sqrt{x} \in \text{Ư}(9)$

$\Rightarrow \sqrt{x} \in \{\pm 1; \pm 3; \pm 9\}$

Vì $x > 0$ nên $\sqrt{x} > 0$

$\Rightarrow \sqrt{x} \in \{1; 3; 9\}$

$\Rightarrow x \in \{1; 9; 81\}$ (thỏa mãn đkxđ)