Question

1/1X3 + 1/3x5 + 1/5x7.......+1/2003x2005 =?

Bạn nào biết đáp án thì cho mk cả cách làm rõ ràng nhé!!!!!

 

Answer 1

\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{2003.2007}\)

\(=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2003}-\frac{1}{2005}\right)\)

\(=\frac{1}{2}\left(1-\frac{1}{2005}\right)\)

\(=\frac{1}{2}.\frac{2004}{2005}\)

\(=\frac{1002}{2005}\)

Answer 2

\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{2003.2005}\)

\(=\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2003}-\frac{1}{2005}\right)\)

\(=\frac{1}{2}.\left(1-\frac{1}{2005}\right)\)

\(=\frac{1}{2}.\frac{2004}{2005}\)

\(=\frac{1002}{2005}\)

Answer 3

Ta có:

\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{2003.2005}\)

\(=\frac{1}{2}.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2003}-\frac{1}{2005}\right)\)

\(=\frac{1}{2}.\left(\frac{1}{1}-\frac{1}{2005}\right)=\frac{1}{2}.\frac{2004}{2005}\)

\(=\frac{1002}{2005}\)

Answer 4

\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+....+\frac{1}{2003.2005}\)

\(=\frac{1}{2}\cdot\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2003}-\frac{1}{2005}\right)\)

\(=\frac{1}{2}\cdot\left(1-\frac{1}{2005}\right)\)

\(=\frac{1}{2}\cdot\frac{2004}{2005}=\frac{1002}{2005}\)

Answer 5

\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{2003.2005}\)

\(=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{2003}-\frac{1}{2005}\right)\)

\(=\frac{1}{2}\left(1-\frac{1}{2005}\right)\)

\(=\frac{1}{2}.\frac{2004}{2005}\)

\(=\frac{1002}{2005}\)