Question

[Minh Ngọc]

a) \(\dfrac{3}{1.4}\) +\(\dfrac{3}{4.7}\) + \(\dfrac{3}{7.10}\) + ... + \(\dfrac{3}{121.124}\)

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

c) \(\dfrac{1}{1.5}\) + \(\dfrac{1}{5.9}\) + \(\dfrac{1}{9.13}\) + ... + \(\dfrac{1}{401.405}\)

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

Answer 1

[Nguyễn Lê Phước Thịnh]

a: \(=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{121}-\dfrac{1}{124}=1-\dfrac{1}{124}=\dfrac{123}{124}\)

b: \(=3\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{100}-\dfrac{1}{101}\right)=3\cdot\dfrac{99}{202}=\dfrac{297}{202}\)

c: \(=\dfrac{1}{4}\left(1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-...+\dfrac{1}{401}-\dfrac{1}{405}\right)=\dfrac{1}{4}\cdot\dfrac{404}{405}=\dfrac{101}{405}\)

d: \(=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{99}-\dfrac{1}{101}=1-\dfrac{1}{101}=\dfrac{100}{101}\)

Answer 2

[Kudo Shinichi AKIRA^_^]

đề bài là j