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

Ta có: \(\hept{\begin{cases}|x+\frac{2006}{2007}|\ge0;\forall x,y\\|\frac{2008}{2009}-y|\ge0;\forall x,y\end{cases}}\)\(\Rightarrow|x+\frac{2006}{2007}|+|\frac{2008}{2009}-y|\ge0;\forall x,y\)

Do đó : \(\Rightarrow|x+\frac{2006}{2007}|+|\frac{2008}{2009}-y|=0\)

\(\Leftrightarrow\hept{\begin{cases}|x+\frac{2006}{2007}|=0\\|\frac{2008}{2009}-y|=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=\frac{-2006}{2007}\\y=\frac{2008}{2009}\end{cases}}\)

Vậy ...

quá đơn giản

n = 1

0 x {2008 x 2009 - 2006 x 2007} = 0

vậy 2009 x 1 - 2009 =0

{2009 x n - 2009} : {2008 x 2009 - 2006 x 2007} = 0  => {2009 x n - 2009} = 0 => n= 1

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 \(A>B\).

để được tổng =0 thì x + 2006/2007 = 0 và 2008/2009 - y =0

vậy suy ra x + 2006/2007 = 0 ; x = -2006/2007

suy ra 2008/2009 - y = 0 ; y = 2008/2009

Vì \(\left|x+\frac{2006}{2007}\right|\ge0;\left|\frac{2008}{2009}-y\right|\ge0\)

Mà \(\left|x+\frac{2006}{2007}\right|+\left|\frac{2008}{2009}-y\right|=0\)

=> \(\hept{\begin{cases}\left|x+\frac{2006}{2007}\right|=0\\\left|\frac{2008}{2009}-y\right|=0\end{cases}}\)=> \(\hept{\begin{cases}x+\frac{2006}{2007}=0\\\frac{2008}{2009}-y=0\end{cases}}\)=> \(\hept{\begin{cases}x=-\frac{2006}{2007}\\y=\frac{2008}{2009}\end{cases}}\)