Question

Tính tổng

a) \(\dfrac{2}{1.3}\)+ \(\dfrac{2}{3.5}\)+ \(\dfrac{2}{5.7}\)+......+\(\dfrac{2}{99.101}\)

b) \(\dfrac{2}{1.3}\)+ \(\dfrac{5}{3.5}\)+ \(\dfrac{5}{5.7}\)+.......+ \(\dfrac{5}{99.101}\)

Answer 1

Trả lời

a)\(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...\dfrac{2}{99.101}\)

=\(2.\left(\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{99.101}\right)\)

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

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

=\(2.\dfrac{100}{101}\)

=\(\dfrac{200}{101}\)

Answer 2

Hình như phần b bạn chép đề sai hay sao đấy

Answer 3

a/\(\dfrac{2}{1.3}+\dfrac{2}{3.5}+...+\dfrac{2}{99.101}\)

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

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

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

=\(\dfrac{2}{2}.\dfrac{100}{101}\)

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

Bài b mk k bít làm!Chúc bn hc tốt!

Answer 4

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

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

A=1.\(\dfrac{100}{101}\)

A=\(\dfrac{100}{101}\)

Answer 5

a) \(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+...+\dfrac{2}{99\cdot101}\)

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

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

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

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

b) \(\dfrac{5}{1\cdot3}+\dfrac{5}{3\cdot5}+\dfrac{5}{5\cdot7}+...+\dfrac{5}{99\cdot101}\)

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

\(=\dfrac{5}{2}\left(1-\dfrac{1}{101}\right)\)

\(=\dfrac{5}{2}\left(\dfrac{101}{101}-\dfrac{1}{101}\right)\)

\(=\dfrac{5}{2}\cdot\dfrac{100}{101}\)

\(=\dfrac{250}{101}\)