a) Ta có:

\(-\dfrac{24}{35}< -\dfrac{24}{30}< -\dfrac{19}{30}\)

\(\Rightarrow x< y\)

b) Ta có:

\(A=\dfrac{2006}{2007}-\dfrac{2007}{2008}+\dfrac{2008}{2009}-\dfrac{2009}{2010}\)

\(A=\left(1-\dfrac{1}{2007}\right)-\left(1-\dfrac{1}{2008}\right)+\left(1-\dfrac{1}{2009}\right)-\left(1-\dfrac{1}{2010}\right)\)

\(A=1-\dfrac{1}{2007}-1+\dfrac{1}{2008}+1-\dfrac{1}{2009}-1+\dfrac{1}{2010}\)

\(A=-\dfrac{1}{2007}+\dfrac{1}{2008}-\dfrac{1}{2009}+\dfrac{1}{2010}\)

Ta lại có:

\(B=-\dfrac{1}{2006.2007}-\dfrac{1}{2008.2009}\)

\(B=-\dfrac{1}{2006}+\dfrac{1}{2007}-\dfrac{1}{2008}+\dfrac{1}{2009}\)

=> Dễ dàng thấy A > B

Khó quá bạn ơi !!!

Đợi mk nghĩ chút nha.

hjhj

\(A=\dfrac{2006}{2007}-\dfrac{2007}{2008}+\dfrac{2008}{2009}-\dfrac{2009}{2010}\)

\(A=\left(1-\dfrac{1}{2007}\right)-\left(1-\dfrac{1}{2008}\right)+\left(1-\dfrac{1}{2009}\right)-\left(1-\dfrac{1}{2010}\right)\)

\(A=1-\dfrac{1}{2007}-1+\dfrac{1}{2008}+1-\dfrac{1}{2009}-1+\dfrac{1}{2010}\)

\(A=\left(1-1\right)+\left(1-1\right)-\dfrac{1}{2007}+\dfrac{1}{2008}-\dfrac{1}{2009}+\dfrac{1}{2010}\)

\(A=\dfrac{1}{2007}+\dfrac{1}{2008}-\dfrac{1}{2009}+\dfrac{1}{2010}\)

\(B=-\dfrac{1}{2006.2007}-\dfrac{1}{2008.2009}\)

\(B=-\left(\dfrac{1}{2006}-\dfrac{1}{2007}\right)-\left(\dfrac{1}{2008}-\dfrac{1}{2009}\right)\)

\(B=-\dfrac{1}{2006}+\dfrac{1}{2007}-\dfrac{1}{2008}+\dfrac{1}{2009}\)

\(B=\dfrac{1}{2007}+\dfrac{1}{2009}-\dfrac{1}{2006}+\dfrac{1}{2008}\)

Dễ dàng thấy \(A>B\)

\(B=1+\left(\frac{2007}{2}+1\right)+\left(\frac{2006}{3}+1\right)+...+\left(\frac{1}{2008}+1\right)=2009\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2008}+\frac{1}{2009}\right)\Rightarrow\frac{A}{B}=\frac{1}{2009}\)

\(B=\dfrac{2008}{1}+\dfrac{2007}{2}+\dfrac{2006}{3}+...+\dfrac{2}{2007}+\dfrac{1}{2008}\)

\(B=1+\left(\dfrac{2007}{2}+1\right)+\left(\dfrac{2006}{3}+1\right)+...+\left(\dfrac{2}{2007}+1\right)+\left(\dfrac{1}{2008}+1\right)\)

\(B=\dfrac{2009}{2009}+\dfrac{2009}{2}+\dfrac{2009}{3}+..+\dfrac{2009}{2007}+\dfrac{2009}{2008}\)

\(B=2009\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2007}+\dfrac{1}{2008}+\dfrac{1}{2009}\right)\)

\(\dfrac{A}{B}=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2007}+\dfrac{1}{2008}+\dfrac{1}{2009}}{2009\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2007}+\dfrac{1}{2008}+\dfrac{1}{2009}\right)}\)

\(\dfrac{A}{B}=\dfrac{1}{2009}\)

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

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

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

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

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

Ta có: \(\left\{{}\begin{matrix}\frac{1}{2007}< \frac{1}{2006}\\\frac{1}{2008}< \frac{1}{2006}\\\frac{1}{2009}< \frac{1}{2006}\end{matrix}\right.\Rightarrow\frac{1}{2007}+\frac{1}{2008}+\frac{1}{2009}< \frac{1}{2006}+\frac{1}{2006}+\frac{1}{2006}=\frac{3}{2006}\)

\(\Rightarrow\frac{1}{2007}+\frac{1}{2008}+\frac{1}{2009}-\frac{3}{2006}< 0\)

\(\Rightarrow4-\left(\frac{1}{2007}+\frac{1}{2008}+\frac{1}{2009}-\frac{3}{2006}\right)>4\)

hay \(A>4\)

\(\text{Vậy A>4}\)

Giải:

Ta có:

A=2009²⁰⁰⁸+1/2009²⁰⁰⁹+1

2009A=2009²⁰⁰⁹+2009/2009²⁰⁰⁹+1

2009A=2009²⁰⁰⁹+1+2008/2009²⁰⁰⁹+1

2009A=2009²⁰⁰⁹+1/2009²⁰⁰⁹+1 + 2008/2009²⁰⁰⁹+1

2009A=1+2008/2009²⁰⁰⁹+1

Tương tự:

B=2009²⁰⁰⁹+1/2009²⁰¹⁰+1

2009B=1+2008/2009²⁰¹⁰+1

Vì 2008/2009²⁰⁰⁹+1 > 2008/2009²⁰¹⁰+1 nên 2009A>2009B

⇒A>B