Ta co [tex]\sqrt[3]{6\sqrt{3}+10}-\sqrt[3]{6\sqrt{3}+10}= \sqrt[3]{(1+\sqrt{3})^{3}}-\sqrt[3]{(\sqrt{3}-1)^{3}}[/tex]
= [tex]\sqrt{3}+1 - \sqrt{3} +1[/tex] = 2
Vậy .....................
Ta có: [tex](\sqrt{2007}+\sqrt{2009})^{2} = 2007+2009+ 2\sqrt{2007+2009} = 2.2008 +2\sqrt{(4016)}[/tex]= 2.2008 + [tex]2\sqrt{2.2008}[/tex]
[tex](2\sqrt{2008})^{2}= 4 . 2008[/tex] = 2.2018 + 2. [tex]\sqrt{2008.2008}[/tex]
Ta thấy : [tex]2\sqrt{2008.2008}> 2\sqrt{2.2008}[/tex]
[tex]\rightarrow (\sqrt{2007}+\sqrt{2009})^{2} < (2\sqrt{2008})^{2} \rightarrow \sqrt{2007}+\sqrt{2009}< 2\sqrt{2008}[/tex]