Đáp án:
$\dfrac{\sqrt{3}+1}{2}$
Giải thích các bước giải:
$\dfrac{\sqrt{2+\sqrt{3}}}{2}:\left(\dfrac{\sqrt{2+\sqrt{3}}}{2}-\dfrac{2}{\sqrt{6}}+\dfrac{\sqrt{2+\sqrt{3}}}{2\sqrt{3}}\right)\\ =\dfrac{\sqrt{2}\sqrt{2+\sqrt{3}}}{2\sqrt{2}}:\left(\dfrac{\sqrt{2}\sqrt{2+\sqrt{3}}}{2\sqrt{2}}-\dfrac{2}{\sqrt{6}}+\dfrac{\sqrt{2}\sqrt{2+\sqrt{3}}}{\sqrt{2}.2\sqrt{3}}\right)\\ =\dfrac{\sqrt{4+2\sqrt{3}}}{2\sqrt{2}}:\left(\dfrac{\sqrt{4+3\sqrt{3}}}{2\sqrt{2}}-\dfrac{2}{\sqrt{6}}+\dfrac{\sqrt{4+2\sqrt{3}}}{2\sqrt{6}}\right)\\ =\dfrac{\sqrt{3+2\sqrt{3}+1}}{2\sqrt{2}}:\left(\dfrac{\sqrt{3+2\sqrt{3}+1}}{2\sqrt{2}}-\dfrac{2}{\sqrt{6}}+\dfrac{\sqrt{3+2\sqrt{3}+1}}{2\sqrt{6}}\right)\\ =\dfrac{\sqrt{3}+1}{2\sqrt{2}}:\left(\dfrac{\sqrt{3}+1}{2\sqrt{2}}-\dfrac{2}{\sqrt{6}}+\dfrac{\sqrt{3}+1}{2\sqrt{6}}\right)\\ =\dfrac{\sqrt{3}+1}{2\sqrt{2}}:\left(\dfrac{3+\sqrt{3}}{2\sqrt{6}}-\dfrac{4}{2\sqrt{6}}+\dfrac{\sqrt{3}+1}{2\sqrt{6}}\right)\\ =\dfrac{\sqrt{3}+1}{2\sqrt{2}}:\dfrac{3+\sqrt{3}-4+\sqrt{3}+1}{2\sqrt{6}}\\ =\dfrac{\sqrt{3}+1}{2\sqrt{2}}:\dfrac{2\sqrt{3}}{2\sqrt{6}}\\ =\dfrac{\sqrt{3}+1}{2\sqrt{2}}.\dfrac{2\sqrt{6}}{2\sqrt{3}}\\ =\dfrac{\sqrt{3}+1}{2}$
