Có $x + 2\sqrt{2x - 4} = 2 + 2\sqrt{2(x - 2)} + x - 2 = (\sqrt{2})^{2} + 2.\sqrt{2}.\sqrt{x - 2} + (\sqrt{x - 2})^{2} = (\sqrt{2} + \sqrt{x - 2})^{2}$
Tương tự có: $x - 2\sqrt{2x - 4} = 2 - 2\sqrt{2(x - 2)} + x - 2 = (\sqrt{2})^{2} - 2.\sqrt{2}.\sqrt{x - 2} + (\sqrt{x- 2})^{2} = (\sqrt{2} - \sqrt{x - 2})^{2}$
Suy ra: $A = \sqrt{x + 2\sqrt{2x - 4}} + \sqrt{x - 2\sqrt{2x - 4}} = \sqrt{(\sqrt{2} + \sqrt{x - 2})^{2}} +\sqrt{(\sqrt{2} - \sqrt{x - 2})^{2}} = \sqrt{2} + \sqrt{x - 2} + \left | \sqrt{2} - \sqrt{x - 2} \right |$
Nếu $\sqrt{2} > \sqrt{x - 2} \Leftrightarrow 2 < x < 4$ thì $A = \sqrt{2} + \sqrt{x - 2} + \sqrt{x} - \sqrt{x - 2}= 2\sqrt{2}$
Nếu $\sqrt{2} < \sqrt{x - 2} \Leftrightarrow x > 4$ thì $A = \sqrt{2} + \sqrt{x - 2} - \sqrt{x} + \sqrt{x - 2} = 2\sqrt{x - 2}$
