\(\left(\sqrt{2003}+\sqrt{2005}\right)^2=2003+2005+2\sqrt{2003.2005}=4008+2\sqrt{2003.2005}\)
\(\left(2\sqrt{2004}\right)^2=4.2004=2.2004+2.2004=4008+2.2004\)
TA có 2003.2005 = (2004 -1 )(2004 + 1 ) = 2004 ^2 - 1 <2004 ^2
=> 2003 . 2005 < 2004^2 =>\(\sqrt{2003.2005}
Ta có:20042-1<20044
=>2003.2005<20042
=>2\(\sqrt{2003.2005}\)<2.2004
Do 2003+2005=2004+2004
=>2003+2\(\sqrt{2003.2005}\)+2005<2004+2.2004+2004
=>\(\left(\sqrt{2003}+\sqrt{2005}\right)^2
so sanh hai can bac hai so hoc sau
√(6+√6+√6) va 3