Đáp án + Giải thích các bước giải:

Điều kiện: $0 < x \le 4$

$$\frac{2 + \sqrt{x}}{\sqrt{2} + \sqrt{2 + \sqrt{x}}} + \frac{2 - \sqrt{x}}{\sqrt{2} - \sqrt{2 - \sqrt{x}}} = \sqrt{2}$$$$\Leftrightarrow \frac{\sqrt{2}(2 + \sqrt{x})}{2 + \sqrt{4 + 2\sqrt{x}}} + \frac{\sqrt{2}(2 - \sqrt{x})}{2 - \sqrt{4 - 2\sqrt{x}}} = \sqrt{2}$$$$\Leftrightarrow \frac{2 + \sqrt{x}}{2 + \sqrt{(\sqrt{2} + \sqrt{x})^2}} + \frac{2 - \sqrt{x}}{2 - \sqrt{(\sqrt{2} - \sqrt{x})^2}} = 1$$$$\Leftrightarrow \frac{2 + \sqrt{x}}{2 + \sqrt{2} + \sqrt{x}} + \frac{2 - \sqrt{x}}{2 - \vert{}\sqrt{2} - \sqrt{x}\vert{}} = 1$$

• TH1: $0 < x \le 2 \Rightarrow \vert{}\sqrt{2} - \sqrt{x}\vert{} = \sqrt{2} - \sqrt{x}$

$$\frac{2 + \sqrt{x}}{2 + \sqrt{2} + \sqrt{x}} + \frac{2 - \sqrt{x}}{2 - \sqrt{2} + \sqrt{x}} = 1$$$$\Leftrightarrow 6 + 4\sqrt{x} - 2\sqrt{2x} = x + 4\sqrt{x} + 2$$$$\Leftrightarrow x + 2\sqrt{2x} - 4 = 0$$$$\Leftrightarrow (\sqrt{x} + \sqrt{2})^2 = 6$$$$\Leftrightarrow \sqrt{x} = \sqrt{6} - \sqrt{2}$$$$\Leftrightarrow x = 8 - 4\sqrt{3} \quad (\text{thỏa mãn})$$

• TH2: $2 < x \le 4 \Rightarrow \vert{}\sqrt{2} - \sqrt{x}\vert{} = \sqrt{x} - \sqrt{2}$

$$\frac{2 + \sqrt{x}}{2 + \sqrt{2} + \sqrt{x}} + \frac{2 - \sqrt{x}}{2 + \sqrt{2} - \sqrt{x}} = 1$$$$\Leftrightarrow 8 + 4\sqrt{2} - 2x = 6 + 4\sqrt{2} - x$$$$\Leftrightarrow x = 2 \quad (\text{loại})$$

Kết luận: $x = 8 - 4\sqrt{3}$