\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{97.99}\)
\(=\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}\right)\)
\(=\frac{1}{3}.\left(1-\frac{1}{99}\right)\)
\(=\frac{1}{3}.\frac{98}{99}\)
\(=\frac{98}{297}\)
Chuc bn học tốt
\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{97.99}\)
\(=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{99}\)
\(=1-\frac{1}{99}\)
\(=\frac{98}{99}\)
Đặt tổng là M
Ta có
\(M=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+....+\frac{1}{97}-\frac{1}{99}\right)\)
\(\Rightarrow M=\frac{1}{2}\left(1-\frac{1}{99}\right)\)
\(\Rightarrow M=\frac{1}{2}.\frac{98}{99}=\frac{49}{99}\)
Đặt A=\(\frac{1}{1.3}\)+\(\frac{1}{3.5}+\frac{1}{5.7}+............+\frac{1}{97.99}\)
2A=\(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+....................+\frac{1}{97}-\frac{1}{99}\)
2A=1-\(\frac{1}{99}\)
2A=\(\frac{98}{99}\)
A=\(\frac{49}{99}\)
\(\frac{1}{1.3}\) + \(\frac{1}{3.5}\) + \(\frac{1}{5.7}\) + ... +\(\frac{1}{99.100}\) = \(\frac{1}{2}\) . ( 1- \(\frac{1}{3}\) + \(\frac{1}{3}\) - \(\frac{1}{4}\) + \(\frac{1}{5}\) -\(\frac{1}{7}\) +.... + \(\frac{1}{97}\) - \(\frac{1}{99}\) ) = \(\frac{1}{2}\) . ( 1 - \(\frac{1}{99}\) ) = \(\frac{1}{2}\). \(\frac{98}{99}\) = \(\frac{98}{198}\) bạn tự rút gọn nha
Đặt A=\(\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+...+\frac{1}{97\cdot99}\)
\(2A=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{97}-\frac{1}{99}\)
\(2A=1-\frac{1}{99}\)
\(2A=\frac{98}{99}\)
\(A=\frac{98}{99}:2\)
\(A=\frac{49}{99}\)
