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a) ĐKXĐ: $a \ge 0$, $b \ge 0$, $a \ne b$, $a \ne 1$.

$P = \left( \frac{3\sqrt{a}}{a + \sqrt{ab} + b} - \frac{3a}{a\sqrt{a} - b\sqrt{b}} + \frac{1}{\sqrt{a} - \sqrt{b}} \right) : \frac{(a - 1)(\sqrt{a} - \sqrt{b})}{2a + 2\sqrt{ab} + 2b}$

$= \frac{3\sqrt{a}(\sqrt{a} - \sqrt{b}) - 3a + (a + \sqrt{ab} + b)}{(\sqrt{a} - \sqrt{b})(a + \sqrt{ab} + b)} \cdot \frac{2(a + \sqrt{ab} + b)}{(a - 1)(\sqrt{a} - \sqrt{b})}$

$= \frac{(\sqrt{a} - \sqrt{b})^2}{(\sqrt{a} - \sqrt{b})(a + \sqrt{ab} + b)} \cdot \frac{2(a + \sqrt{ab} + b)}{(a - 1)(\sqrt{a} - \sqrt{b})} = \frac{2}{a - 1}$

b) $P \in \mathbb{Z} \implies (a - 1) \in \text{Ư}(2) = \{-2; -1; 1; 2\}$

$a - 1 = -2 \implies a = -1$ (loại)

$a - 1 = -1 \implies a = 0$ (thỏa mãn)

$a - 1 = 1 \implies a = 2$ (thỏa mãn)

$a - 1 = 2 \implies a = 3$ (thỏa mãn)

Vậy ....