Ta có :\(\dfrac{2008.2009+4018}{2010.2011-4020}=\dfrac{2008.2009+2009.2}{2010.2011-2010.2}\)

\(=\dfrac{2009.\left(2008+2\right)}{2010\left(2011-2\right)}=\dfrac{2009.2010}{2010.2009}=1\)

Dễ thấy:

\(\frac{2008}{2009}>\frac{2008}{2009+2010+2011}\)

\(\frac{2009}{2010}>\frac{2009}{2009+2010+2011}\)

\(\frac{2010}{2011}>\frac{2010}{2009+2010+2011}\)

=>\(\frac{2008}{2009}+\frac{2009}{2010}+\frac{2010}{2011}>\frac{2008}{2009+2010+2011}+\frac{2009}{2009+2010+2011}+\frac{2010}{2009+2010+2011}\)

Hay \(\frac{2008}{2009}+\frac{2009}{2010}+\frac{2010}{2011}>\frac{2008+2009+2010}{2009+2010+2011}\)

Vậy A > B

A = \(\dfrac{2008}{2009+2010+2011}+\dfrac{2009}{2009+2010+2011}+\dfrac{2010}{2009+2010+2011}\)

Ta có: 

\(\dfrac{2008}{2009}>\dfrac{2008}{2009+2010+2011}\)

\(\dfrac{2009}{2010}>\dfrac{2009}{2009+2010+2011}\)

\(\dfrac{2010}{2011}>\dfrac{2010}{2009+2010+2011}\)

Từ 3 điều trên suy ra : A < B

mình cũng có bài giống như này nhưng chưa làm được

mình cũng thế

Tớ cũng có bài này nhưng chưa làm được

cau tra loi la 50 khong can biet lam the nao

Bạn chỉ cần lấy : (2008/2009+2009/2010+2010/2011+2011/2008)-4=số dương

vậy (2008+...2008) > 4