\(=\frac{1}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+....+\frac{2}{2003.2005}\right)\)
=\(\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+.....+\frac{1}{2003}-\frac{1}{2005}\right)\)
=\(\frac{1}{2}\left(1-\frac{1}{2005}\right)=\frac{1}{2}.\frac{2004}{2005}=\frac{1002}{2005}\)
\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+....+\frac{1}{2003.2005}=\)
\(=\frac{2}{2.1.3}+\frac{2}{2.3.5}+\frac{2}{2.5.7}+....+\frac{2}{2.2003.2005}\)
\(=\frac{1}{2}.\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+....+\frac{1}{2003.2005}\right)\)
\(=\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+....+\frac{1}{2003}-\frac{1}{2005}\right)\)
\(=\frac{1}{2}.\left(1-\frac{1}{2005}\right)\)
\(=\frac{1}{2}.\frac{2004}{2005}\)
\(=\frac{1002}{2005}\)
Chúc bạn học tốt nha!
\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{2003.2005}\)
\(\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{2003.2005}\right)\)
\(\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2003}-\frac{1}{2005}\right)\)
\(\frac{1}{2}.\left(1-\frac{1}{2005}\right)\)
\(\frac{1}{2}.\frac{2004}{2005}\)
\(\frac{2004}{2.2005}=\frac{1002}{2005}\)
= 2/1.3.2 + 2/3.5.2 + 2/5.7.2 + ...+ 2/2003.2005.2
= 1/2 . ( 2/1.3 + 2/3.5 + 2/5.7 +...+ 2/2003.2005)
= 1/2 . ( 1/1- 1/3 + 1/3 - 1/5 + 1/5 - 1/7 +...+ 1/2003 - 1/2005)
= 1/2 . (1/1 - 1/2005)
= 1/2. 2004/2005
= 1002/2005
đạt A=1/1x3 + 1/3x5 +........+1/2003x2005
=>2A=2/1x3 + 2/3x5 +......+2/2003x2005
=>2A=(1-1/3)+(1/3-1/5)+....+(1/2003-1/2005)
=>2A=1-1/3+1/3-1/5+1/5-1/7+.....+1/2003-1/2005
=>2A=1-1/2005
=>A=1002/2005
ai thấy đúng thì kik mik nha
