Rút gọn biểu thức : G = 1 + căn bậc hai của 3/2/1 + căn bậc hai của 1 + căn bậc hai của 3/2

$\begin{array}{l}G=\frac{1+\frac{\sqrt{3}}{2}}{1+\sqrt{1+\frac{\sqrt{3}}{2}}}+\frac{1-\frac{\sqrt{3}}{2}}{1-\sqrt{1-\frac{\sqrt{3}}{2}}}=\frac{\frac{2+\sqrt{3}}{2}}{\frac{\sqrt{2}+\sqrt{2+\sqrt{3}}}{\sqrt{2}}}+\frac{\frac{2-\sqrt{3}}{2}}{\frac{\sqrt{2}-\sqrt{2-\sqrt{3}}}{\sqrt{2}}}\\ =\frac{2+\sqrt{3}}{2}.\frac{\sqrt{2}}{\sqrt{2}+\sqrt{2+\sqrt{3}}}+\frac{2-\sqrt{3}}{2}.\frac{\sqrt{2}}{\sqrt{2}-\sqrt{2-\sqrt{3}}}\end{array}$

$\begin{array}{l}=\frac{2+\sqrt{3}}{\sqrt{2}.\left(\sqrt{2}+\sqrt{2+\sqrt{3}}\right)}+\frac{2-\sqrt{3}}{\sqrt{2}.\left(\sqrt{2}-\sqrt{2-\sqrt{3}}\right)}\\ =\frac{2+\sqrt{3}}{2+\sqrt{4+2\sqrt{3}}}+\frac{2-\sqrt{3}}{2-\sqrt{4-2\sqrt{3}}}=\frac{2+\sqrt{3}}{2+\sqrt{3}+1}+\frac{2-\sqrt{3}}{2-\left(\sqrt{3}-1\right)}\\ =\frac{2+\sqrt{3}}{3+\sqrt{3}}+\frac{2-\sqrt{3}}{3-\sqrt{3}}=\frac{\left(2+\sqrt{3}\right)\left(3-\sqrt{3}\right)+\left(2-\sqrt{3}\right)\left(3+\sqrt{3}\right)}{9-3}\\ =\frac{6-3+\sqrt{3}+6-3-\sqrt{3}}{6}=\frac{6}{6}=1\end{array}$