A = \(\sqrt{7+3\sqrt{5}}+\sqrt{7-3\sqrt{5}}\)
<=> A2 = ( \(\sqrt{7+3\sqrt{5}}+\sqrt{7-3\sqrt{5}}\) )2
= 7 + \(3\sqrt{5}\) + \(2\sqrt{\left(7+3\sqrt{5}\right).\left(7-3\sqrt{5}\right)}\) + 7 - \(3\sqrt{5}\)
= 14 + 2\(\sqrt{7^2-\left(3\sqrt{5}\right)^2}\)
= 14 + 2\(\sqrt{4}\)
= 18
=> A = \(\sqrt{18}\)
B = \(\sqrt{2-\sqrt{2\sqrt{5}-2}}\) - \(\sqrt{2+\sqrt{2\sqrt{5}-2}}\)
<=> B2 = ( \(\sqrt{2-\sqrt{2\sqrt{5}-2}}\) - \(\sqrt{2+\sqrt{2\sqrt{5}-2}}\) )2
= 2 - \(\sqrt{2-2\sqrt{5}}\) - 2\(\sqrt{\left(2-\sqrt{2\sqrt{5}-2}\right)\left(2+\sqrt{2\sqrt{5}-2}\right)}\) + 2 + \(\sqrt{2-2\sqrt{5}}\)
= 4 - 2 \(\sqrt{2^2-\left(\sqrt{2\sqrt{5}-2}\right)^2}\)
= 4 - 2 \(\sqrt{4-\left(2\sqrt{5}-2\right)}\)
= 4 - 2 \(\sqrt{4-2\sqrt{5}+2}\)
= 4 - 2 \(\sqrt{\left(\sqrt{5}-1\right)^2}\)
= 4 - 2( \(\sqrt{5}-1\) )
= 6 - 2\(\sqrt{5}\)
=> B = \(\sqrt{6-2\sqrt{5}}\) = \(\sqrt{\left(\sqrt{5}-1\right)^2}\) = \(\sqrt{5}-1\)
