Question

BT1: Tính

1) A= \(\dfrac{1}{2.4}+\dfrac{1}{4.6}+\dfrac{1}{6.8}+...+\dfrac{1}{20.22}\)

Answer 1

A=\(\dfrac{1}{2\cdot4}+\dfrac{1}{4\cdot6}+\dfrac{1}{6\cdot8}+...+\dfrac{1}{20\cdot22}\)

\(=\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{8}+...+\dfrac{1}{20}-\dfrac{1}{22}\right)\)

\(=\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{22}\right)\)

\(=\dfrac{1}{2}\left(\dfrac{11}{22}-\dfrac{1}{22}\right)\)

\(=\dfrac{1}{2}\cdot\dfrac{5}{11}\)

\(=\dfrac{5}{22}\)

Answer 2

Mk có thể làm wen đc k