Question

Tính tỉ số \(\dfrac{A}{B}\), biết:

\(A=\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2007}+\dfrac{1}{2008}+\dfrac{1}{2009}\\ B=\dfrac{2008}{1}+\dfrac{2007}{2}+\dfrac{2006}{3}+...+\dfrac{2}{2007}+\dfrac{1}{2008}\)

Answer 1

Có: \(B=\dfrac{2008}{1}+\dfrac{2007}{2}+...+\dfrac{1}{2008}\)

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

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

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

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

 

Answer 2

Ta có:

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

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

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

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

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