Question

All giúp mk với 

Answer 1

\(A=\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{49.50}\)

\(=\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{49}-\dfrac{1}{50}\)

\(=1-\dfrac{1}{50}\)

\(=\dfrac{49}{50}\)

Answer 2

\(A=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{49}-\dfrac{1}{50}\)

\(=1-\dfrac{1}{50}=\dfrac{49}{50}\)

Answer 3

Answer 4

A = \(\dfrac{ 1}{ 1 . 2 } + \dfrac{1}{ 2 . 3 } + ....... + \dfrac{1}{ 49 . 50 }\)

A = \(1 - \dfrac{1}{2} + \dfrac{1 }{2} - \dfrac{1}{ 3 } + \dfrac{1}{3} + ...... - \dfrac{1}{49} + \dfrac{1}{49} - \dfrac{1}{50} \)  

A = \(1 - \dfrac{1}{50}\)

A = \(\dfrac{ 50}{50} - \dfrac{1}{50}\)

A = \(\dfrac{49}{50}\) 

Vậy \(A = \dfrac{49}{50}\)