Đặt \(A=\left(\sqrt{2+\sqrt{3}}-\sqrt{3+\sqrt{5}}\right)^2\)
Ta có: \(4A=\left(2\sqrt{2+\sqrt{3}}-2\sqrt{3+\sqrt{5}}\right)^2\)
\(=\left(\sqrt{4+2\sqrt{3}}-\sqrt{6+2\sqrt{5}}\right)^2\)
\(=\left(\sqrt{\left(\sqrt{3}+1\right)^2}-\sqrt{\left(\sqrt{5}+1\right)^2}\right)^2\)
\(=\left(\sqrt{3}+1-\sqrt{5}-1\right)^2=\cdot\left(\sqrt{3}-\sqrt{5}\right)^2\)
\(=8-2\sqrt{15}\)
\(\Rightarrow A=2-\frac{\sqrt{15}}{2}\)