Question

[Jenny Jenny]

Tính A= \(\dfrac{2}{1.2}+\dfrac{2}{2.3}+.......+\dfrac{2}{98.99}+\dfrac{2}{99.100}\)

Answer 1

[Huy Thắng Nguyễn]

A = \(\dfrac{2}{1\times2}+\dfrac{2}{2\times3}+...+\dfrac{2}{98\times99}+\dfrac{2}{99\times100}\)

A = \(2\left(\dfrac{1}{1\times2}+\dfrac{1}{2\times3}+...+\dfrac{1}{98\times99}+\dfrac{1}{99\times100}\right)\)

= \(2\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{98}-\dfrac{1}{99}+\dfrac{1}{99}-\dfrac{1}{100}\right)\)

= \(2\left(1-\dfrac{1}{100}\right)\)

= \(2\times\dfrac{99}{100}=\dfrac{99}{50}\)

Answer 2

[Nguyễn Phương Mai]

có cách giải khác vẫn ra đc kết quả đúng