\(A=\sqrt{3+\sqrt{5-\sqrt{\left(2\sqrt{3}+1\right)^2}}}=\sqrt{3+\sqrt{5-\left(2\sqrt{3}+1\right)}}\)
\(=\sqrt{3+\sqrt{4-2\sqrt{3}}}=\sqrt{3+\sqrt{\left(\sqrt{3}-1\right)^2}}\)
\(=\sqrt{3+\sqrt{3}-1}=\sqrt{2+\sqrt{3}}=\dfrac{1}{\sqrt{2}}\sqrt{4+2\sqrt{3}}\)
\(=\dfrac{1}{\sqrt{2}}\sqrt{\left(\sqrt{3}+1\right)^2}=\dfrac{\sqrt{3}+1}{\sqrt{2}}=\dfrac{\sqrt{2}+\sqrt{6}}{2}\)
\(A=\sqrt{3+\sqrt{5-\sqrt{13+\sqrt{48}}}}\\ =\sqrt{3+\sqrt{5-\sqrt{\left(1+2\sqrt{3}\right)^2}}}\\ =\sqrt{3+\sqrt{5-1+2\sqrt{3}}}\\ =\sqrt{3+\sqrt{\left(\sqrt{3}-1\right)^2}}\\ =\sqrt{3+\sqrt{3}-1}\\ =\sqrt{2+\sqrt{3}}\)
Ta có: \(A=\sqrt{3+\sqrt{5-\sqrt{13+\sqrt{48}}}}\)
\(=\sqrt{3+\sqrt{4-2\sqrt{3}}}\)
\(=\sqrt{3+\sqrt{3}-1}\)
\(=\sqrt{2-\sqrt{3}}\)
\(=\dfrac{\sqrt{4-2\sqrt{3}}}{\sqrt{2}}\)
\(=\dfrac{\sqrt{6}-\sqrt{2}}{2}\)
