\(G=\sqrt{3+\sqrt{5}}+\sqrt{7-3\sqrt{5}}-\sqrt{2}\)
\(=\sqrt{\frac{1}{2}\left(6+2\sqrt{5}\right)}+\sqrt{\frac{1}{2}\left(14-6\sqrt{5}\right)}-\sqrt{2}\)
\(=\sqrt{\frac{1}{2}}\left(\sqrt{1+2\cdot1\cdot\sqrt{5}+5}+\sqrt{9-2\cdot3\cdot\sqrt{5}+5}\right)-\sqrt{2}\)
\(=\frac{\sqrt{2}}{2}\left(\sqrt{\left(1+\sqrt{5}\right)^2}+\sqrt{\left(3-\sqrt{5}\right)^2}\right)-\frac{\sqrt{2}}{2}\cdot2\)
\(=\frac{\sqrt{2}}{2}\left(\left|1+\sqrt{5}\right|+\left|3-\sqrt{5}\right|-2\right)\)
\(=\frac{\sqrt{2}}{2}\left(1+\sqrt{5}+3-\sqrt{5}-2\right)\)
\(=\frac{\sqrt{2}}{2}\cdot2=\sqrt{2}\)
Vậy \(G=\sqrt{2}\).
