a) 8 và [tex]\sqrt{15}+\sqrt{17}[/tex]. Ta có:
[tex]8^2=64=32+2\sqrt{256}\\ (\sqrt{15}+\sqrt{17})^2=32+2\sqrt{15.17}=32+2\sqrt{(16-1)(16+1)}< 32+2\sqrt{256}\\ \Rightarrow 8> \sqrt{15}+\sqrt{17}[/tex]
b) Ta có:
[tex]2=1+1<1+\sqrt{2}\Rightarrow 2< 1+\sqrt{2}[/tex]
c) Ta có:
[tex](\sqrt{2007}+\sqrt{2009})^2=4016+2\sqrt{2007.2009}(*)\\ (2\sqrt{2008})^2=4.2008=4016+2\sqrt{2008.2008}>4016+2\sqrt{(2008-1)(2008+1)}=4016+2\sqrt{2007.2009}\\ \Rightarrow \sqrt{2007}+\sqrt{2009}<2\sqrt{2008}[/tex]