[tex]\sqrt{6+2\sqrt{2}+2\sqrt{3}+2\sqrt{6}}=\sqrt{(\sqrt{2}+\sqrt{3}+1)^2}=\sqrt{2}+\sqrt{3}+1[/tex]
b) [tex]\sqrt{10+\sqrt{60}+\sqrt{20}+\sqrt{40}}=\sqrt{10+2\sqrt{15}+2\sqrt{5}+2\sqrt{10}}=\sqrt{(\sqrt{2}+\sqrt{3}+\sqrt{5})^2}=\sqrt{2}+\sqrt{3}+\sqrt{5}[/tex]
c)[tex]\sqrt{6-2\sqrt{2}+\sqrt{12}+\sqrt{18-\sqrt{128}}}=\sqrt{6-2\sqrt{2}+\sqrt{12}+\sqrt{18-8\sqrt{2}}}=\sqrt{6-2\sqrt{2}+2\sqrt{3}+\sqrt{(4-\sqrt{2})^2}}=\sqrt{10-3\sqrt{2}+2\sqrt{3}}[/tex] đề sai không nhỉ