Cách khác: B=\(\left(3-\sqrt{5}\right)\sqrt{3+\sqrt{5}}+\left(3+\sqrt{5}\right)\sqrt{3-\sqrt{5}}\)
<=>\(\sqrt{2}B=\left(3-\sqrt{5}\right)\sqrt{2}.\sqrt{3+\sqrt{5}}+\left(3+\sqrt{5}\right)\sqrt{2}.\sqrt{3-\sqrt{5}}\)
<=> \(\sqrt{2}B=\left(3-\sqrt{5}\right)\sqrt{6+2\sqrt{5}}+\left(3+\sqrt{5}\right)\sqrt{6-2\sqrt{5}}\)
<=> \(\sqrt{2}B=\left(3-\sqrt{5}\right)\sqrt{\left(\sqrt{5}+1\right)^2}+\left(3+\sqrt{5}\right)\sqrt{\left(\sqrt{5}-1\right)^2}\)
<=>\(\sqrt{2}B=\left(3-\sqrt{5}\right)\left(\sqrt{5}+1\right)+\left(3+\sqrt{5}\right)\left|\sqrt{5}-1\right|\)
=\(3\sqrt{5}+3-5-\sqrt{5}+\left(3+\sqrt{5}\right)\left(\sqrt{5}-1\right)\)
= \(3\sqrt{5}+3-5-\sqrt{5}+3\sqrt{5}-3+5-\sqrt{5}\)
= \(6\sqrt{5}-2\sqrt{5}\)
= \(4\sqrt{5}\)
=> B= \(\frac{4\sqrt{5}}{\sqrt{2}}=2\sqrt{10}\)
