Question

\(\dfrac{3}{1.2}+\dfrac{3}{2.3}+\dfrac{3}{3.4}+...+\dfrac{3}{100.101}\)=?

Answer 1

=> 3.( \(\dfrac{1}{1.2}\)+\(\dfrac{1}{2.3}\)+\(\dfrac{1}{3.4}\)+...+\(\dfrac{1}{100.101}\))

=> 3.(\(\dfrac{1}{1}\)-\(\dfrac{1}{2}\)+\(\dfrac{1}{2}\)-\(\dfrac{1}{3}\)+...+\(\dfrac{1}{100}\)-\(\dfrac{1}{101}\))

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

=> 3. \(\dfrac{100}{101}\)

=> \(\dfrac{300}{101}\)

Tick cho mk nhé, chúc bạn học tốt

Answer 2

\(\dfrac{3}{1.2}+\dfrac{3}{2.3}+\dfrac{3}{3.4}+...+\dfrac{3}{100.101}\)

= \(3.\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{100.101}\right)\)

= \(3.\left(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...\dfrac{1}{100}-\dfrac{1}{101}\right)\).

= \(3.\left(1-\dfrac{1}{101}\right)\)= \(3.\dfrac{100}{101}=\dfrac{300}{101}\).

Answer 3

\(\dfrac{3}{1.2}+\dfrac{3}{2.3}+\dfrac{3}{3.4}+...+\dfrac{3}{100.101}\)

\(=3\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{100.101}\right)\)

\(=3\left(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{100}-\dfrac{1}{101}\right)\)

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

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

\(=3.\dfrac{100}{101}\)

\(=\dfrac{300}{101}\)

Vậy \(\dfrac{3}{1.2}+\dfrac{3}{2.3}+\dfrac{3}{3.4}+...+\dfrac{3}{100.101}=\dfrac{300}{101}\)

Answer 4

\(\dfrac{3}{1.2}+\dfrac{3}{2.3}+...+\dfrac{3}{100.101}\)

\(=3\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{100.101}\right)\)

\(=3\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{100}-\dfrac{1}{101}\right)\)

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

=\(3\times\dfrac{100}{101}\)

=\(\dfrac{300}{101}\)