\(\sqrt{9-3\sqrt{5}}-\sqrt{9+3\sqrt{5}}=\dfrac{1}{\sqrt{2}}\left(\sqrt{18-6\sqrt{5}}-\sqrt{18+6\sqrt{5}}\right)\)
\(=\dfrac{1}{\sqrt{2}}\left(\sqrt{\left(\sqrt{15}-\sqrt{3}\right)^2}-\sqrt{\left(\sqrt{15}+\sqrt{3}\right)^2}\right)\)
\(=\dfrac{1}{\sqrt{2}}\left(\sqrt{15}-\sqrt{3}-\sqrt{15}-\sqrt{3}\right)=-\dfrac{2\sqrt{3}}{\sqrt{2}}=-\sqrt{6}\)
\(\sqrt{4-\sqrt{15}}+\sqrt{4+\sqrt{15}}-2\sqrt{3-\sqrt{5}}\)
\(=\dfrac{1}{\sqrt{2}}\left(\sqrt{8-2\sqrt{15}}+\sqrt{8+2\sqrt{15}}-2\sqrt{6-2\sqrt{5}}\right)\)
\(=\dfrac{1}{\sqrt{2}}\left(\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}+\sqrt{\left(\sqrt{5}+\sqrt{3}\right)^2}-2\sqrt{\left(\sqrt{5}-1\right)^2}\right)\)
\(=\dfrac{1}{\sqrt{2}}\left(\sqrt{5}-\sqrt{3}+\sqrt{5}+\sqrt{3}-2\left(\sqrt{5}-1\right)\right)\)
\(=\dfrac{2}{\sqrt{2}}=\sqrt{2}\)
k: Ta có: \(\sqrt{9-3\sqrt{5}}-\sqrt{9+3\sqrt{5}}\)
\(=\dfrac{\sqrt{18-6\sqrt{5}}-\sqrt{18+6\sqrt{5}}}{\sqrt{2}}\)
\(=\dfrac{\sqrt{15}-\sqrt{3}-\sqrt{15}-\sqrt{3}}{\sqrt{2}}\)
\(=-\sqrt{6}\)
