Ta có công thức: \(1+n^2+\frac{n^2}{\left(n+1\right)^2}=\left(1+n-\frac{n}{n+1}\right)^2\)

Chứng minh công thức: \(\left(1+n-\frac{n}{n+1}\right)^2\)

\(=1^2+n^2+\frac{n^2}{\left(n+1\right)^2}+2n-\frac{2n}{n+1}-\frac{2n^2}{n+1}\)

\(=1+n^2+\frac{n^2-2n^2\left(n+1\right)+2n\left(n+1\right)^2-2n\left(n+1\right)}{\left(n+1\right)^2}\)

\(=1+n^2+\frac{n^2-2n^3-2n^2+2n\left(n^2+2n+1\right)-2n^2-2n}{\left(n+1\right)^2}\)

\(=1+n^2+\frac{-2n^3-3n^2-2n+2n^3+4n^2+2n}{\left(n+1\right)^2}=1+n^2+\frac{n^2}{\left(n+1\right)^2}\)

\(A=\sqrt{1+2008^2+\frac{2008^2}{2009^2}}+\frac{2008}{2009}\)

\(=1+2008-\frac{2008}{2009}+\frac{2008}{2009}\)

=2009

=>A là số tự nhiên

sun rang roi ha ha ha ha ha 

ko biết làm

đặt \(2008=a\)

\(\sqrt{1+a^2+\frac{a^2}{\left(a+1\right)^2}}=\sqrt{\left(a+1\right)^2-\frac{2\left(a+1\right).a}{a+1}+\left(\frac{a}{a+1}\right)^2}=\)\(\sqrt{\left(a+1-\frac{a}{a+1}\right)^2}=a+1-\frac{a}{a+1}\)=2008+1- \(\frac{2008}{2009}\)

=> A= 2008+1 = 2009

ta có A = 2008^2009+2 / 2008^2009-1 = 2008^2009-1+3 / 2008^2009-1 = 1 + 3/2008^2009-1

lại có B = 2008^2009 / 2008^2009-3 = 2008^2009-3+3 / 2008^2009-3 = 1 + 3/2008^2009-3

vì 3/2008^2009-1 < 3/2008^2009-3 => 1 + 3/2008^2009-1 < 1 + 3/2008^2009-3

Hay A<B 

Vậy A<B

^-^

vế trái = \(\dfrac{2008+1}{\sqrt{2008}}+\dfrac{2009-1}{\sqrt{2009}}=\sqrt{2008}+\sqrt{2009}+\dfrac{1}{\sqrt{2008}}-\dfrac{1}{\sqrt{2009}}\)

vì \(\dfrac{1}{\sqrt{2008}}-\dfrac{1}{\sqrt{2009}}>0\) nên suy ra đpcm