A>b

Cách làm: Bạn tách |B ra rồi so sánh với từng ps ở A, sau đó Kết luận

\(A=\frac{2006}{2007}+\frac{2007}{2008}+\frac{2008}{2009}=1-\frac{1}{2007}+1-\frac{1}{2008}+1-\frac{1}{2009}\)

\(=3-\frac{1}{2007}-\frac{1}{2008}-\frac{1}{2009}>1\).

\(B=\frac{2006+2007+2008}{2007+2008+2009}< \frac{2007+2008+2009}{2007+2008+2009}=1\).

Suy ra \(A>B\).

Ta có:
\(x=\dfrac{2006}{2007}-\dfrac{2007}{2008}+\dfrac{2008}{2009}-\dfrac{2009}{2010}\)

\(=\dfrac{2006.2008-2007^2}{2007.2008}+\dfrac{2008.2010-2009^2}{2009.2010}\)

\(=\dfrac{2006.2007+2006-2007^2}{2007.2008}+\dfrac{2008.2009+2008-2009^2}{2009.2010}\)

\(=\dfrac{2007\left(2006-2007\right)+2006}{2007.2008}+\dfrac{2009\left(2008-2009\right)+2008}{2009.2010}\)

\(=\dfrac{-1}{2007.2008}+\dfrac{-1}{2008.2010}< \dfrac{-1}{2006.2007}+\dfrac{1}{2007.2008}\)

\(\Rightarrow x< y\)

Vậy x < y

2006/2007<1

2007/2008<1

2008<2009<1

2009/2006>1

A=2006/2007+2007/2008+2008/2009+2009/2006\(\approx\)3+1=4

 \(A\approx4\)

Ta có:

2006/2007 + 2007/2008 + 2008/2009 + 2009/2006

= 1 - 1/2007 + 1 - 1/2008 + 1 - 1/2009 + 1 + 3/2006

= (1 + 1 + 1 + 1) - (1/2007 + 1/2008 + 1/2009) + 3/2006

= 4 - (1/2007 + 1/2008 + 1/2009) + 3/2006

Vì 1/2007 < 1/2006

1/2008 < 1/2006

1/2009 < 1/2006

=> 1/2007 + 1/2008 + 1/2009 < 3/2006

=> -(1/2007 + 1/2008 + 1/2009) + 3/2006 > 0

=> 4 - (1/2007 + 1/2008 + 1/2009) + 3/2006 > 4 - 0 = 4

=> 2006/2007 + 2007/2008 + 2008/2009 + 2009/2006 > 4

Ta có:

2006/2007 + 2007/2008 + 2008/2009 + 2009/2006

= 1 - 1/2007 + 1 - 1/2008 + 1 - 1/2009 + 1 + 3/2006

= (1 + 1 + 1 + 1) - (1/2007 + 1/2008 + 1/2009) + 3/2006

= 4 - (1/2007 + 1/2008 + 1/2009) + 3/2006

Vì 1/2007 < 1/2006

1/2008 < 1/2006

1/2009 < 1/2006

=> 1/2007 + 1/2008 + 1/2009 < 3/2006

=> -(1/2007 + 1/2008 + 1/2009) + 3/2006 > 0

=> 4 - (1/2007 + 1/2008 + 1/2009) + 3/2006 > 4 - 0 = 4

=> 2006/2007 + 2007/2008 + 2008/2009 + 2009/2006 > 4

hghghhbjbhghdbvhvvdvdvdvdvdvv vdvdv

Ta có : \(A=\frac{2006}{2007}+\frac{2007}{2008}+\frac{2008}{2009}+\frac{2009}{2006}\)

                \(=\frac{2007-1}{2007}+\frac{2008-1}{2008}+\frac{2009-1}{2009}+\frac{2006+3}{2006}\)

                  \(=1-\frac{1}{2007}+1-\frac{1}{2008}+1-\frac{1}{2009}+1+\frac{3}{2006}\)

                  \(=\left(1+1+1+1\right)-\left(\frac{1}{2007}+\frac{1}{2008}+\frac{1}{2009}-\frac{3}{2006}\right)\)

                  \(< 4-\left(\frac{1}{2009}+\frac{1}{2009}+\frac{1}{2009}-\frac{3}{2009}\right)\)     

                    \(=4\)

=> A < 4 

Vậy A < 4 

Ta có A= (1 -1/2007) +(1-1/2008)+(1-1/2009)+(1+3/2006)= 4-(1/2007+1/2008+1/2009-3/2006) <4