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![](https://rs.olm.vn/images/avt/0.png?1311)
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
![](https://rs.olm.vn/images/avt/0.png?1311)
Vì 20009 x 2009 + 2008 < 2009 x 2009 + 2009
=>A < 1
Ta có: \(B=\frac{2009x2009+2009}{2008x2009+2010}=\frac{2009x\left(2008+1\right)+2009}{2008x2009+2010}=\frac{2008x2009+2009+2009}{2008x2009+2010}\)
\(B=\frac{2008x2009+4018}{2008x2009+2010}=\frac{2008x2009+2010+2008}{2008x2009+2010}=\frac{2008x2009+2010}{2008x2009+2010}+\frac{2008}{2008x2009+2010}\)
\(B=1+\frac{2008}{2008x2009+2010}>1\)
Mà A < 1
=>A < B
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![](https://rs.olm.vn/images/avt/0.png?1311)
\(b,S=\frac{2007}{2008}+\frac{2008}{2009}+\frac{2009}{2010}+\frac{2010}{2011}\)
\(\text{Ta có: }\frac{2007}{2008}< 1\)
\(\frac{2008}{2009}< 1\)
\(\frac{2009}{2010}< 1\)
\(\frac{2010}{2011}< 1\)
\(\Rightarrow\frac{2007}{2008}+\frac{2008}{2009}+\frac{2009}{2010}+\frac{2010}{2011}< 1+1+1+1\)
\(\Rightarrow\frac{2007}{2008}+\frac{2008}{2009}+\frac{2009}{2010}+\frac{2010}{2011}< 4\)
Ta có : b = 2009 x 2019
b = 2009 x (2018+1) = 2009 x 2018 + 2009
Ta lại có : a = 2008 x 2010
a = 2008 x (2009 + 1) = 2008 x 2009 + 2008
Ta thấy :2009 x 2018 + 2009 > 2008 x 2009 + 2008
Vậy b>a
Ta có :
a = 2008 x 2010 = 2008 x ( 2009 + 1 ) = 2008 x 2009 + 2008
b = 2009 x 2009 = 2009 x ( 2008 + 1 ) = 2009 x 2008 + 2009
Do 2008 x 2009 + 2008 < 2009 x 2008 + 2009
=> a < b
Vậy a < b
Chúc bạn học tốt !!!