Question

1/1.3+1/3.5+1/5.7+....1/2003.2005

Answer 1

tính tổng 

Answer 2

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

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

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

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

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

Answer 3

yêu Sáng

Answer 4

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

\(=\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{2003.2005}\right)\)

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

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

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

Answer 5

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

\(2A=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{2003.2005}\)

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

\(2A=1-\frac{1}{2005}\)

\(2A=\frac{2004}{2005}\)

\(A=\frac{1007}{2005}\)

Answer 6

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

\(=\frac{1.2}{2.1.3}+\frac{1.2}{2.3.5}+\frac{1.2}{2.5.7}+...+\frac{1}{2.2003.2005}\)

\(=\frac{1}{2}.\frac{2}{1.3}+\frac{1}{2}.\frac{2}{3.5}+\frac{1}{2}.\frac{2}{5.7}+...+\frac{1}{2}.\frac{2}{2003.2005}\)

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

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

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

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