Question

 

Bài toán 2. Tính tỉ số 

 biết:

 

Answer 1

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

Answer 2

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

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

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

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

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

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

=>B=2009.A

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