\(\frac{2}{3.5}+\frac{2}{5.7}+.................+\frac{2}{97.99}\)
=\(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+..................+\frac{1}{97}-\frac{1}{99}\)
=\(\frac{1}{3}-\frac{1}{99}\)
=\(\frac{32}{99}\)
\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+.....+\frac{1}{97}-\frac{1}{99}\)
\(=\frac{1}{3}-\frac{1}{99}\)
\(=\frac{32}{99}\)
Ta có: \(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{97.99}\)
\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{97}-\frac{1}{99}\)
\(=\frac{1}{3}-\frac{1}{99}\)
\(=\frac{32}{99}\)
\(\frac{5-3}{5\cdot3}+\frac{7-5}{5\cdot7}+..........+\frac{99-97}{97\cdot99}\)
\(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+..........+\frac{1}{97}-\frac{1}{99}\)
\(\Rightarrow A=\frac{1}{3}-\frac{1}{99}\Rightarrow A=\frac{32}{99}\)
Ghi lại đè
\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+....+\frac{1}{97}-\frac{1}{99}\)
\(=\frac{1}{3}+\left(-\frac{1}{5}\right)+\frac{1}{5}+\left(-\frac{1}{7}\right)+\frac{1}{7}+\left(-\frac{1}{9}\right)+\frac{1}{9}+...+\left(-\frac{1}{97}\right)+\frac{1}{97}+-\frac{1}{99}\)
\(=\frac{1}{3}+\left(-\frac{1}{5}+\frac{1}{5}\right)+\left(-\frac{1}{7}+\frac{1}{7}\right)+\left(-\frac{1}{9}+\frac{1}{9}\right)+...+\left(-\frac{1}{97}+\frac{1}{97}\right)+\left(-\frac{1}{99}\right)\)
\(=\frac{1}{3}+0+0+0+...+0+\left(-\frac{1}{99}\right)\)
\(=\frac{1}{3}+\left(-\frac{1}{99}\right)\)
\(=\frac{34}{99}\)
hết fim
lộn kết quả nha thât ra là \(\frac{32}{99}\)