Đặt $A=\dfrac{\dfrac{\sqrt{2+\sqrt3}}{2}}{\dfrac{\sqrt{2+\sqrt3}}{2}-\dfrac{2}{\sqrt6}+\dfrac{\sqrt{2+\sqrt3}}{2\sqrt3}}$
Nhân tử số và mẫu số với $2\sqrt2$ ta có:
$A=\dfrac{\sqrt{4+2\sqrt3}}{\sqrt{4+2\sqrt3}-\dfrac{4\sqrt2}{\sqrt6}+\dfrac{\sqrt{4+2\sqrt3}}{\sqrt3}}$
Ta có :$\sqrt{4+2\sqrt3}=\sqrt{3+2\sqrt3+1}\\=\sqrt{(\sqrt3+1)^2}=\sqrt3+1$
$\Rightarrow A=\dfrac{\sqrt3+1}{\sqrt3+1-\dfrac{4\sqrt3}{3}+\dfrac{\sqrt3+3}{3}}\\=\dfrac{\sqrt3+1}{\dfrac{3\sqrt3+3-4\sqrt3+\sqrt3+3}{3}}\\=\dfrac{3(\sqrt3+1)}{4\sqrt3+6-4\sqrt3}\\=\dfrac{3(\sqrt3+1)}{6}\\=\dfrac{\sqrt3+1}{2}$
