Question

\(\dfrac{4}{1.4}+\dfrac{4}{5.9}+....+\dfrac{4}{2001.2005}\)

Answer 1

$\dfrac{4}{1.4}+\dfrac{5.9}+....+\dfrac{4}{2001.2005}$
$=1+\dfrac15-\dfrac19+....+\dfrac{1}{2001}-\dfrac{1}{2005}$
$=1-\dfrac{1}{2005}=\dfrac{2004}{2005}$

Answer 2

 \(\dfrac{4}{1.4}+\dfrac{4}{5.9}+...+\dfrac{4}{2001.2005}\)

\(=1+\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{13}+...+\dfrac{1}{2001}-\dfrac{1}{2005}\)

\(=1+\dfrac{1}{5}-\dfrac{1}{2005}\)

\(=1+\dfrac{401}{2005}-\dfrac{1}{2005}\)

\(=1+\dfrac{400}{2005}=1+\dfrac{80}{401}=\dfrac{481}{401}\)

Answer 3

Giải:

\(\dfrac{4}{1.4}+\dfrac{4}{5.9}+...+\dfrac{4}{2001.2005}\) 

\(=1+\dfrac{1}{5}-\dfrac{1}{9}+...+\dfrac{1}{2001}-\dfrac{1}{2005}\) 

\(=1+\dfrac{1}{5}-\dfrac{1}{2005}\) 

\(=1+\dfrac{80}{401}\) 

\(=\dfrac{481}{401}\)