Lời giải:
\(A=\sqrt{10+\sqrt{24}+\sqrt{40}+\sqrt{60}}=\sqrt{10+2\sqrt{6}+2\sqrt{10}+2\sqrt{15}}\)
\(=\sqrt{10+2\sqrt{2}(\sqrt{3}+\sqrt{5})+2\sqrt{15}}=\sqrt{2+(3+5+2\sqrt{15})+2\sqrt{2}(\sqrt{3}+\sqrt{5})}\)
\(=\sqrt{2+(\sqrt{3}+\sqrt{5})^2+2\sqrt{2}(\sqrt{3}+\sqrt{5})}\)
\(=\sqrt{(\sqrt{2}+\sqrt{3}+\sqrt{5})^2}=\sqrt{2}+\sqrt{3}+\sqrt{5}\)
\(2B=2.\frac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+4}{\sqrt{2}+\sqrt{3}+\sqrt{4}}=2.\frac{(\sqrt{2}+\sqrt{3}+\sqrt{4})+(\sqrt{4}+\sqrt{6}+\sqrt{8})}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(=2.\frac{(\sqrt{2}+\sqrt{3}+\sqrt{4})+\sqrt{2}(\sqrt{2}+\sqrt{3}+\sqrt{4})}{\sqrt{2}+\sqrt{3}+\sqrt{4}}=2(1+\sqrt{2})\)
Do đó:
\(A-2B=\sqrt{3}+\sqrt{5}-(2+\sqrt{2})>\sqrt{2}+\sqrt{4}-(2+\sqrt{2})=0\)
\(\Rightarrow A>2B\)
