Ta có
$A = \dfrac{(\sqrt{2} + \sqrt{3} + 2) + (2 + \sqrt{6} + \sqrt{8})}{\sqrt{2} + \sqrt{3} + \sqrt{4}}$
$= \dfrac{\sqrt{2} + \sqrt{3} + \sqrt{4}}{\sqrt{2} + \sqrt{3} + \sqrt{4}} + \dfrac{\sqrt{4} + \sqrt{6} + \sqrt{8}}{\sqrt{2} + \sqrt{3} + \sqrt{4}}$
$= 1 + \dfrac{\sqrt{2} (\sqrt{2} + \sqrt{3} + \sqrt{4})}{\sqrt{2} + \sqrt{3} + \sqrt{4}}$
$= 1 + \sqrt{2}$.