\(A=\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+.....+\dfrac{2}{97.99}\)
Ta thấy:\(\dfrac{1}{1}-\dfrac{1}{3}=\dfrac{2}{1.3
}\)
\(\dfrac{1}{3}-\dfrac{1}{5}=\dfrac{2}{3.5}\)
............\(\dfrac{1}{97}-\dfrac{1}{99}=\dfrac{2}{97.99}\)
\(\Rightarrow A=\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+..........+\dfrac{1}{97}-\dfrac{1}{99}\) =\(\dfrac{1}{1}-\dfrac{1}{99}\)
=\(\dfrac{99}{99}-\dfrac{1}{99}\)
=\(\dfrac{98}{99}\)
Vậy A=\(\dfrac{98}{99}\)
A = \(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+...+\dfrac{2}{97.99}\)
A = \(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{97}-\dfrac{1}{99}\)
A = \(1-\dfrac{1}{99}\)
A = \(\dfrac{98}{99}\)
A=\(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{97}-\dfrac{1}{99}\)
A=\(1-\dfrac{1}{99}=\dfrac{98}{99}\)
