\(a)\ B = \dfrac{x - 2}{x + 2\sqrt{x}} - \dfrac{1}{\sqrt{x}} + \dfrac{1}{\sqrt{x} + 2}\ (x > 0)\)

\(\Leftrightarrow B = \dfrac{x - 2}{\sqrt{x}(\sqrt{x} + 2)} - \dfrac{1}{\sqrt{x}} + \dfrac{1}{\sqrt{x} + 2}\)

\(\Leftrightarrow B = \dfrac{x - 2 - (\sqrt{x} + 2) + \sqrt{x}}{\sqrt{x}(\sqrt{x} + 2)}\)

\(\Leftrightarrow B = \dfrac{x - 2 - \sqrt{x} - 2 + \sqrt{x}}{\sqrt{x}(\sqrt{x} + 2)}\)

\(\Leftrightarrow B = \dfrac{x - 4}{\sqrt{x}(\sqrt{x} + 2)}\)

\(\Leftrightarrow B = \dfrac{(\sqrt{x} - 2)(\sqrt{x} + 2)}{\sqrt{x}(\sqrt{x} + 2)}\)

\(\Leftrightarrow B = \dfrac{\sqrt{x} - 2}{\sqrt{x}}\)

\(b)\) Có \(x = 3 - 2\sqrt{2} = (\sqrt{2} - 1)^2\)

\(\Leftrightarrow \sqrt{x} = \sqrt{(\sqrt{2} - 1)^2} = \sqrt{2} - 1\)

Thay vào biểu thức \(B\) ta được:

\(B = \dfrac{(\sqrt{2} - 1) - 2}{\sqrt{2} - 1}\)

\(\Leftrightarrow B = \dfrac{\sqrt{2} - 3}{\sqrt{2} - 1}\)

\(\Leftrightarrow B = \dfrac{(\sqrt{2} - 3)(\sqrt{2} + 1)}{(\sqrt{2} - 1)(\sqrt{2} + 1)}\)

\(\Leftrightarrow B = \dfrac{2 + \sqrt{2} - 3\sqrt{2} - 3}{2 - 1}\)

\(\Leftrightarrow B = -1 - 2\sqrt{2}\)

\(c)\) Ta có \(B = \dfrac{\sqrt{x} - 2}{\sqrt{x}} = 1 - \dfrac{2}{\sqrt{x}}\)

Để \(B \in \mathbb{Z} \rightarrow \dfrac{2}{\sqrt{x}} \in \mathbb{Z}\)

\(\Rightarrow \sqrt{x}\) là ước dương của \(2\)

\(\Rightarrow \sqrt{x} \in \{1; 2\}\)

\(\Rightarrow x \in \{1; 4\}\)

Do \(x \in \mathbb{N},x > 0\Rightarrow x \in \{1;4\}\)