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a)

$P = \left( \frac{x - 2}{\sqrt{x}(\sqrt{x} + 2)} + \frac{\sqrt{x}}{\sqrt{x}(\sqrt{x} + 2)} \right) \cdot \frac{\sqrt{x} + 1}{\sqrt{x} - 1}$

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

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

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

b)

$2P = 2\sqrt{x} + 5 \Leftrightarrow 2 \cdot \frac{\sqrt{x} + 1}{\sqrt{x}} = 2\sqrt{x} + 5$

$\Leftrightarrow \frac{2\sqrt{x} + 2}{\sqrt{x}} = 2\sqrt{x} + 5$

$\Leftrightarrow 2\sqrt{x} + 2 = 2x + 5\sqrt{x}$

$\Leftrightarrow 2x + 3\sqrt{x} - 2 = 0$

$\Leftrightarrow (2\sqrt{x} - 1)(\sqrt{x} + 2) = 0$

$\Leftrightarrow \left[\begin{array}{l} 2\sqrt{x} - 1 = 0 \\ \sqrt{x} + 2 = 0 \text{ (vô nghiệm)} \end{array}\right.$

$\Leftrightarrow \sqrt{x} = \frac{1}{2} \Leftrightarrow x = \frac{1}{4} \text{ (thỏa mãn)}$

Vậy $x = \frac{1}{4}$.