\(C=\frac{x-1}{2009}+\frac{x-2}{2008}=\frac{x-3}{2007}+\frac{x-4}{2006}\)

     \(\Rightarrow\left(\frac{x-1}{2009}-1\right)+\left(\frac{x-2}{2008}-1\right)=\left(\frac{x-3}{2007}-1\right)+\left(\frac{x-4}{2006}-1\right)\)

      \(\Rightarrow\frac{x-1-2009}{2009}+\frac{x-2-2008}{2008}=\frac{x-3-2007}{2007}+\frac{x-4-2006}{2006}\)

      \(\Rightarrow\frac{x-2010}{2009}+\frac{x-2010}{2008}-\frac{x-2010}{2007}-\frac{x-2010}{2006}=0\)

     \(\Rightarrow\left(x-2010\right)\times\left(\frac{1}{2009}+\frac{1}{2008}-\frac{1}{2007}-\frac{1}{2006}\right)=0\)

Vì \(\frac{1}{2009}+\frac{1}{2008}-\frac{1}{2007}-\frac{1}{2006}\ne0\)

Nên x - 2010 = 0

=> x = 2010

Vậy x = 2010

\(\frac{x-1}{2009}+\frac{x-1}{2008}-2=\frac{x-3}{2007}+\frac{x-4}{2006}-2\)

\(\frac{x-2010}{2009}+\frac{x-2010}{2008}-\frac{x-2010}{2007}-\frac{x-2010}{2006}=0\)

\(x-2010\cdot\left(\frac{1}{2009}+\frac{1}{2008}-\frac{1}{2007}-\frac{1}{2006}\right)=0\)

mà vế phải ( vế có phân số ) khác 0

=> x - 2010 = 0

=> x = 2010

Vậy,.........

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:

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

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