a) Ta có: \(\sqrt{3+\sqrt{5}}+\sqrt{3-\sqrt{5}}\)\(=\dfrac{\sqrt{6+2\sqrt{5}}+\sqrt{6-2\sqrt{5}}}{\sqrt{2}}\)\(=\dfrac{\sqrt{5}+1+\sqrt{5}-1}{\sqrt{2}}=\dfrac{2\sqrt{5}}{\sqrt{2}}=\sqrt{10}\)b) Ta có: \(2\sqrt{2}\left(\sqrt{3}-2\right)+\left(9+4\sqrt{2}\right)\)\(=2\sqrt{6}-4\sqrt{2}+9+4\sqrt{2}\)\(=9+2\sqrt{6}\)c) Ta có: \(\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{7+4\sqrt{3}}}}}\)\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\left(2+\sqrt{3}\right)}}}\)\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{28-10\sqrt{3}}}}\)\(=\sqrt{4+\sqrt{5\sqrt{3}+5\left(5-\sqrt{3}\right)}}\)\(=\sqrt{4+\sqrt{25}}=\sqrt{4+5}=3\)(đpcm)Câu trả lời: a) Ta có: \(\sqrt{3+\sqrt{5}}+\sqrt{3-\sqrt{5}}\)\(=\dfrac{\sqrt{6+2\sqrt{5}}+\sqrt{6-2\sqrt{5}}}{\sqrt{2}}\)\(=\dfrac{\sqrt{5}+1+\sqrt{5}-1}{\sqrt{2}}=\dfrac{2\sqrt{5}}{\sqrt{2}}=\sqrt{10}\)b) Ta có: \(2\sqrt{2}\left(\sqrt{3}-2\right)+\left(9+4\sqrt{2}\right)\)\(=2\sqrt{6}-4\sqrt{2}+9+4\sqrt{2}\)\(=9+2\sqrt{6}\)c) Ta có: \(\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{7+4\sqrt{3}}}}}\)\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\left(2+\sqrt{3}\right)}}}\)\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{28-10\sqrt{3}}}}\)\(=\sqrt{4+\sqrt{5\sqrt{3}+5\left(5-\sqrt{3}\right)}}\)\(=\sqrt{4+\sqrt{25}}=\sqrt{4+5}=3\)(đpcm)