Question

tính A

A=\(\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{99.101}\)

giúp mk nha các pạn, mk sẽ tick

Answer 1

Ta có :

\(A=\dfrac{1}{3.5}+\dfrac{1}{5.7}+................+\dfrac{1}{99.101}\)

\(\Rightarrow2A=\dfrac{2}{3.5}+\dfrac{2}{5.7}+.................+\dfrac{2}{99.101}\)

\(\Rightarrow2A=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...............+\dfrac{1}{99}-\dfrac{1}{101}\)

\(\Rightarrow2A=\dfrac{1}{3}-\dfrac{1}{101}\)

\(\Rightarrow2A=\dfrac{98}{303}\)

\(\Rightarrow A=\dfrac{49}{303}\)

Answer 2

\(A=\dfrac{1}{3.5}+\dfrac{1}{5.7}+.....+\dfrac{1}{99.101}\)

\(A=\dfrac{1}{2}\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+.....+\dfrac{1}{99}-\dfrac{1}{101}\right)\)

\(A=\dfrac{1}{2}\left(\dfrac{1}{3}-\dfrac{1}{101}\right)=\dfrac{1}{2}.\dfrac{98}{303}=\dfrac{49}{303}\)