1. $\frac{\sqrt{2} + x}{\sqrt{x} + 3} - \frac{5}{x + \sqrt{x} - 6} - \frac{1}{\sqrt{x} - 2}$ với $x \geq 0; x \neq 4$.
$A = \frac{\left ( \sqrt{2} + x \right )\left ( \sqrt{x} - 2 \right )}{\left ( \sqrt{x} + 3 \right )\left ( \sqrt{x} - 2 \right )} - \frac{5}{\left ( \sqrt{x} + 3 \right )\left ( \sqrt{x} - 2 \right )} - \frac{\sqrt{x} + 3}{\left ( \sqrt{x} + 3 \right )\left ( \sqrt{x} - 2 \right )}$
$A = \frac{x - 4 - 5 - \sqrt{x} - 3}{\left ( \sqrt{x} + 3 \right )\left ( \sqrt{x} - 2 \right )}$
$A = \frac{x - \sqrt{x} - 12}{\left ( \sqrt{x} + 3 \right )\left ( \sqrt{x} - 2 \right )}$
$A = \frac{\left ( \sqrt{x} + 3 \right )\left ( \sqrt{x} - 4 \right )}{\left ( \sqrt{x} + 3 \right )\left ( \sqrt{x} - 2 \right )}$
$A = \frac{\sqrt{x} - 4}{\sqrt{x} - 2}$
2. $x = 6 + 4\sqrt{2}$ thỏa mãn ĐK nên thay vào $A$ ta có:
$A = \frac{\sqrt{6 + 4\sqrt{2}} - 4}{\sqrt{6 + 4\sqrt{2}} - 2} = \frac{\sqrt{4 + 2.2.\sqrt{2} + 2} - 4}{\sqrt{4 + 2.2.\sqrt{2} + 2} - 2} = \frac{\sqrt{\left ( 2 + \sqrt{2} \right )^{2}} - 4}{\sqrt{\left ( 2 + \sqrt{2} \right )^{2}} - 2} = \frac{2 + \sqrt{2} - 4}{2 + \sqrt{2} - 2} = \frac{\sqrt{2} - 2}{\sqrt{2}} = \frac{\sqrt{2}\left ( 1 - \sqrt{2} \right )}{\sqrt{2}} = 1 - \sqrt{2}$
