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

^-^

Không rút gọn đc 2009/2008 đã tối giản

\(\dfrac{2009}{2008}\) là phân số tối giản

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

ta có: \(A=\dfrac{2008^{2009}+2}{2008^{2009}-1}=\dfrac{2008^{2009}-1+3}{2008^{2009}-1}=1+\dfrac{3}{2008^{2009}-1}\)

B=\(\dfrac{2008^{2009}}{2008^{2009}-3}=\dfrac{2008^{2009}-3+3}{2008^{2009}-3}=1+\dfrac{3}{2008^{2009}-3}\)

ta thấy: \(1+\dfrac{3}{2008^{2009}-1}\)<\(1+\dfrac{3}{2008^{2009}-3}\)

vậy A<B

a) \(\dfrac{2}{5}+\dfrac{4}{5}\times\dfrac{5}{2}\)

\(=\dfrac{2}{5}+\dfrac{4\times5}{5\times2}\)

\(=\dfrac{2}{5}+\dfrac{4}{2}\)

\(=\dfrac{2}{5}+2\)

\(=\dfrac{2}{5}+\dfrac{10}{5}\)

\(=\dfrac{12}{5}\)

b) \(\dfrac{2008}{2009}-\dfrac{2009}{2008}+\dfrac{1}{2009}+\dfrac{2007}{2008}\)

\(=\left(1-\dfrac{1}{2009}\right)-\left(1+\dfrac{1}{2008}\right)+\dfrac{1}{2009}+\left(1-\dfrac{1}{2008}\right)\)

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

\(=\left(1-1+1\right)-\left(\dfrac{1}{2009}-\dfrac{1}{2009}\right)-\left(\dfrac{1}{2008}+\dfrac{1}{2008}\right)\)

\(=1-\dfrac{2}{2008}\)

\(=\dfrac{2008}{2008}-\dfrac{2}{2008}\)

\(=\dfrac{2006}{2008}\)

\(=\dfrac{1003}{1004}\)

a: =2/5+4/2

=2/5+2

=12/5

b: \(=1-\dfrac{1}{2009}-1-\dfrac{1}{2008}+\dfrac{1}{2009}+1-\dfrac{1}{2008}\)

\(=1-\dfrac{2}{2008}=1-\dfrac{1}{1004}=\dfrac{1003}{1004}\)

`A=\sqrt{1+2008^2+2008^2/2009^2}+2008/2009`

`=\sqrt{1+2008^2+2.2008+2008^2/2009^2-2.2008}+2008/2009`

`=\sqrt{(2008+1)^2-2.2008+2008^2/2009^2}+2008/2009`

`=\sqrt{2009-2.2008/2009*2009+2008^2/2009^2}+2008/2009`

`=\sqrt{(2009-2008/2009)^2}+2008/2009`

`=|2009-2008/2009|+2008/2009`

`=2009-2008/2009+2008/2009`

`=2009` là 1 số tự nhiên