$a)$ 

$Q = \frac{\sqrt{x} + 1}{\sqrt{x} - 2}$ với $x \ge 0, x \ne 4$

Thay $x = 25$ (TMĐK) vào $Q$, ta được:

$Q = \frac{\sqrt{25} + 1}{\sqrt{25} - 2}$

$= \frac{5 + 1}{5 - 2}$

$= \frac{6}{3}$

$= 2$

$b)$ 

$P = \frac{\sqrt{x} + 1}{\sqrt{x} - 2} + \frac{\sqrt{x}}{\sqrt{x} + 2} - \frac{3\sqrt{x} - 2}{4 - x}$ với $x \ge 0, x \ne 4$

$= \frac{\sqrt{x} + 1}{\sqrt{x} - 2} + \frac{\sqrt{x}}{\sqrt{x} + 2} + \frac{3\sqrt{x} - 2}{x - 4}$

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

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

$= \frac{2x + 4\sqrt{x}}{(\sqrt{x} - 2)(\sqrt{x} + 2)}$

$= \frac{2\sqrt{x}(\sqrt{x} + 2)}{(\sqrt{x} - 2)(\sqrt{x} + 2)}$

$= \frac{2\sqrt{x}}{\sqrt{x} - 2}$ (đpcm)

$c)$ 

$P = \frac{2\sqrt{x}}{\sqrt{x} - 2}$, $Q = \frac{\sqrt{x} + 1}{\sqrt{x} - 2}$ với $x \ge 0, x \ne 4$

$M = P : Q$

$M = \frac{2\sqrt{x}}{\sqrt{x} - 2} : \frac{\sqrt{x} + 1}{\sqrt{x} - 2}$

$= \frac{2\sqrt{x}}{\sqrt{x} - 2} \cdot \frac{\sqrt{x} - 2}{\sqrt{x} + 1}$

$= \frac{2\sqrt{x}}{\sqrt{x} + 1}$

$= \frac{2(\sqrt{x} + 1) - 2}{\sqrt{x} + 1}$

$= 2 - \frac{2}{\sqrt{x} + 1}$

Vì $x \ge 0 \implies \sqrt{x} \ge 0 \implies \sqrt{x} + 1 \ge 1$

$\to 0 < \frac{2}{\sqrt{x} + 1} \le 2$

$\to 0 \le 2 - \frac{2}{\sqrt{x} + 1} < 2$

$\to 0 \le M < 2$

Để $M \in \mathbb{Z}$ thì $M \in \{0; 1\}$

TH1: $M = 0$

$\frac{2\sqrt{x}}{\sqrt{x} + 1} = 0$

$2\sqrt{x} = 0$

$x = 0 \quad (\text{TM})$

TH2: $M = 1$

$\frac{2\sqrt{x}}{\sqrt{x} + 1} = 1$

$2\sqrt{x} = \sqrt{x} + 1$

$\sqrt{x} = 1$

$x = 1 \quad (\text{TM})$

Vậy $x \in \{0; 1\}$ thì $M \in \mathbb{Z}$