\(\sqrt{2003}\)+\(\sqrt{2005}\)<2\(\sqrt{2004}\)
ta có :\(\left(\sqrt{2005}+\sqrt{2003}\right)^2\le\left(1^2+1^2\right)\left(2005+2003\right)=2.4008\)(bđt bu-nhia-cop xki)
\(\left(2\sqrt{2004}\right)^2=4.2004=2.4008\)
→\(\sqrt{2003}+\sqrt{2005}< 2\sqrt{2004}\)
ta có: A=\(\left(\sqrt{2003}+\sqrt{2005}\right)^2=2003+2005+2\sqrt{2003.2005}\)
=\(4008+2\sqrt{2003.2005}\)
=4008+\(2\sqrt{2003.2004+2003}\)(1)
B=\(\left(2\sqrt{2004}\right)^2=8016=4008+2.2004\)
=\(4008+2\sqrt{2004^2}\)
=4008+\(2\sqrt{2003.2004+2004}\)(2)
từ (1) và (2) ==>A<B
