Question

Tính:

\(A=\dfrac{1}{3.5}+\dfrac{1}{5.7}+\dfrac{1}{7.9}+...........+\dfrac{1}{99.101}\)

Answer 1

\(A=\dfrac{1}{3.5}+\dfrac{1}{5.7}+\dfrac{1}{7.9}+............+\dfrac{1}{99.101}\)

\(A=\dfrac{1}{2}\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+.....+\dfrac{1}{99}-\dfrac{1}{101}\right)\)

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

Answer 2

\(A=\dfrac{1}{3.5}+\dfrac{1}{5.7}+\dfrac{1}{7.9}+.................+\dfrac{1}{99.101}\)

\(=\dfrac{1}{2}\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{99}-\dfrac{1}{101}\right)\)

\(=\dfrac{1}{2}\left(\dfrac{1}{3}-\dfrac{1}{101}\right)\)

\(=\dfrac{1}{2}.\dfrac{98}{303}\)

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

Answer 3

\(A=\dfrac{1}{3.5}+\dfrac{1}{5.7}+\dfrac{1}{7.9}+.........+\dfrac{1}{99.101}\)

\(\Rightarrow2A=\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+............+\dfrac{2}{99.101}\)

\(\Rightarrow2A=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+....+\dfrac{1}{99}-\dfrac{1}{101}\)

(do \(\dfrac{n}{a\left(a+n\right)}=\dfrac{1}{a}-\dfrac{1}{a+n}\) với mọi giá trị của \(a\in N\)*)

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

\(\Rightarrow2A=\dfrac{98}{303}\Rightarrow A=\dfrac{49}{303}\)

Chúc bạn học tốt!!!

Answer 4

Ta có

\(A=\dfrac{1}{3.5}+\dfrac{1}{5.7}+\dfrac{1}{7.9}+...+\dfrac{1}{99.101}\)

\(2A=\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+...+\dfrac{2}{99.101}\)

\(2A=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{99}-\dfrac{1}{101}\)

\(2A=\dfrac{1}{3}-\dfrac{1}{101}\)

\(A=\dfrac{98}{303}:2=\dfrac{49}{303}\)

Answer 5

A = \(\dfrac{1}{2}\) . ( \(\dfrac{2}{3.5}\) + \(\dfrac{2}{5.7}\) + \(\dfrac{2}{7.9}\) + ........... + \(\dfrac{2}{99.101}\) )

A = \(\dfrac{1}{2}\) . ( \(\dfrac{1}{3}\) - \(\dfrac{1}{5}\) + \(\dfrac{1}{5}\) - \(\dfrac{1}{7}\) + \(\dfrac{1}{7}\) - \(\dfrac{1}{9}\) + ............ + \(\dfrac{1}{99}\) - \(\dfrac{1}{101}\) )

A = \(\dfrac{1}{2}\) . ( \(\dfrac{1}{3}\) - \(\dfrac{1}{101}\) )

A = \(\dfrac{1}{2}\) . \(\dfrac{98}{303}\)

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

Vậy A = \(\dfrac{49}{303}\)

Answer 6

Giải:

\(A=\dfrac{1}{3.5}+\dfrac{1}{5.7}+\dfrac{1}{7.9}+...+\dfrac{1}{99.101}.\)

\(A=\dfrac{1}{2}\left(\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+...+\dfrac{2}{99.101}\right).\)

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

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

\(A=\dfrac{1}{2}\left(\dfrac{1}{3}+0+0+...+0-\dfrac{1}{101}\right).\)

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

\(A=\dfrac{1}{2}.\dfrac{98}{303}.\)

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

Vậy \(A=\dfrac{49}{303}.\)

~ Học tốt!!! ~