\(A=\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+\frac{1}{10.13}+\frac{1}{13.16}+\frac{1}{16.19}\)

\(A=\frac{1}{3}\cdot\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+....+\frac{1}{16}-\frac{1}{19}\right)\)

\(A=\frac{1}{3}\cdot\left(1-\frac{1}{19}\right)\)

\(A=\frac{1}{3}\cdot\frac{18}{19}=\frac{6}{19}\)

\(B=\frac{1}{32}+\frac{1}{96}+\frac{1}{192}+\frac{1}{320}+\frac{1}{480}\)

\(B=\frac{1}{4\cdot8}+\frac{1}{8\cdot12}+\frac{1}{12\cdot16}+\frac{1}{16\cdot20}+\frac{1}{20\cdot24}\)

\(B=\frac{1}{4}\cdot\left(\frac{1}{4}-\frac{1}{8}+\frac{1}{8}-\frac{1}{12}+\frac{1}{12}-\frac{1}{16}+\frac{1}{16}-\frac{1}{20}+\frac{1}{20}-\frac{1}{24}\right)\)

\(B=\frac{1}{4}\cdot\left(\frac{1}{4}-\frac{1}{24}\right)\)

\(B=\frac{1}{4}\cdot\frac{5}{24}=\frac{5}{96}\)

\(A=\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+\frac{1}{10.13}+\frac{1}{13.16}+\frac{1}{16.19}\)

\(A=\frac{1}{3}\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{16}-\frac{1}{19}\right)\)

\(A=\frac{1}{3}\left(1-\frac{1}{19}\right)\)

\(A=\frac{1}{3}.\frac{18}{19}\)

\(A=\frac{6}{19}\)

\(B=\frac{1}{32}+\frac{1}{96}+\frac{1}{192}+\frac{1}{320}+\frac{1}{480}\)

\(B=\frac{1}{4.8}+\frac{1}{8.12}+\frac{1}{12.16}+\frac{1}{16.20}+\frac{1}{20.24}\)

\(B=\frac{1}{4}\left(\frac{1}{4}-\frac{1}{8}+\frac{1}{8}-\frac{1}{12}+...+\frac{1}{20}-\frac{1}{24}\right)\)

\(B=\frac{1}{2}\left(\frac{1}{4}-\frac{1}{24}\right)\)

\(B=\frac{1}{2}.\frac{5}{24}\)

\(B=\frac{5}{48}\)

\(A=\frac{1}{3}\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{16}-\frac{1}{19}\right)\)

\(A=\frac{1}{3}\left(1-\frac{1}{19}\right)=\frac{1}{3}.\frac{18}{19}=\frac{6}{19}\) VẬY: \(A=\frac{6}{19}\)

\(B=\frac{1}{4.2.4}+\frac{1}{4.4.6}+...+\)\(\frac{1}{4.10.12}\)

\(B=\frac{1}{4}\left(\frac{1}{2.4}+\frac{1}{4.6}+...+\frac{1}{10.12}\right)\)\(\Rightarrow B=\frac{1}{8}\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}+...+\frac{1}{10}-\frac{1}{12}\right)\)

                                                                               \(\Rightarrow B=\frac{1}{8}\left(\frac{1}{2}-\frac{1}{120}\right)=\frac{1}{8}.\frac{59}{120}=\frac{59}{960}\)

Ta thấy: 1/1-1/4 = 3/4 = 3.(1/1.4)

           1/4-1/7 = 3/28 = 3.(1/4.7)

A = 3(1/1-1/4+1/4-1/7+...+1/97-1/100)

A = 3.(1-1/100)

A = 3.(99/100)

A = 297/100

\(A=\frac{1}{3}.\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+.......+\frac{1}{97}-\frac{1}{100}\right)\)

\(A=\frac{1}{3}.\left(1-\frac{1}{100}\right)\)

\(A=\frac{1}{3}.\frac{99}{100}\)

\(A=\frac{33}{100}\)

Bài toán gì mà có cả câu trả lời thế này ????????

Đăng lên mà trả lời luôn thế này thì đăng lên làm gì cho nó mệt

B = \(\frac{1}{4.7}+\frac{1}{7.10}+\frac{1}{10.13}+...+\frac{1}{301.304}\)

B = \(\frac{1}{3}.\left(\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+...+\frac{1}{301}-\frac{1}{304}\right)\)

B = \(\frac{1}{3}.\left(\frac{1}{4}-\frac{1}{304}\right)\)

B = \(\frac{1}{3}.\frac{75}{304}\)

B = \(\frac{25}{304}\)

\(B=\left(\frac{3}{4.7}+\frac{3}{7.10}+\frac{3}{10.13}+...+\frac{3}{301.304}\right):3\)

\(\Rightarrow B=\left(\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{301}-\frac{1}{304}\right):3\)

\(\Rightarrow B=\left(\frac{1}{4}-\frac{1}{304}\right):3\)

\(\Rightarrow B=\frac{75}{304}:3=\frac{25}{304}\)

1. Tìm x

\(\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{97.100}=x\)

\(\Rightarrow\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{97}-\frac{1}{100}=x\)

\(\Rightarrow1-\frac{1}{100}=x\)

\(\Rightarrow x=\frac{99}{100}\)

\(2.Tính\)

\(\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}+\frac{1}{110}+\frac{1}{132}\)

\(=\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}+\frac{1}{10.11}+\frac{1}{11.12}\)

\(=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}+\frac{1}{11}-\frac{1}{12}\)

\(=\frac{1}{5}-\frac{1}{12}=\frac{7}{60}\)

học vui!!

Xin lỗi nha. Bài 1 mk làm sai. Lại nè:

\(\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{97.100}=x\)

\(\frac{1}{3}\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{97.100}\right)=x\)

\(\frac{1}{3}.\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{100}\right)=x\)

\(\frac{1}{3}.\left(1-\frac{1}{100}\right)=x\)

\(\frac{1}{3}\cdot\frac{99}{100}=x\)

\(\frac{33}{100}=x\)

1. Tìm x

\(\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+\frac{1}{10.13}+...+\frac{1}{97.100}=x\)

\(\frac{1}{3}.\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+\frac{3}{10.13}+...+\frac{3}{97.100}\right)=x\)

\(\frac{1}{3}.\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+...+\frac{1}{97}-\frac{1}{100}\right)=x\)

\(\frac{1}{3}.\left(1-\frac{1}{100}\right)=x\)

\(\frac{1}{3}.\frac{99}{100}=x\)

\(\frac{1.99}{3.100}=\frac{1.33}{1.100}=x\)

\(\frac{33}{100}=x\)

Vậy :  \(x=\frac{33}{100}\)

2 . tính

\(\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}+\frac{1}{110}+\frac{1}{132}\)

\(=\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}+\frac{1}{10.11}+\frac{1}{11.12}\)

\(=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}+\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}\)

\(=\frac{1}{5}-\frac{1}{12}=\frac{1}{5}+\frac{-1}{12}\)

\(=\frac{12}{60}+\frac{-5}{60}\)

\(=\frac{7}{60}\)

\(A=\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+\frac{1}{7\cdot9}+...+\frac{1}{99\cdot101}\)

\(A=\frac{1}{2}\cdot\left(\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+...+\frac{2}{99\cdot101}\right)\)

\(A=\frac{1}{2}\cdot\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{99}-\frac{1}{101}\right)\)

\(A=\frac{1}{2}\cdot\left(\frac{1}{3}-\frac{1}{101}\right)=\frac{1}{2}\cdot\frac{97}{303}=\frac{97}{606}\)

\(B=\frac{2}{4\cdot7}+\frac{2}{7\cdot10}+\frac{2}{10\cdot13}+...+\frac{2}{100\cdot103}\)

\(B=\frac{2}{3}\cdot\left(\frac{3}{4\cdot7}+\frac{3}{7\cdot10}+\frac{3}{10\cdot13}+...+\frac{3}{100\cdot103}\right)\)

\(B=\frac{2}{3}\cdot\left(\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+...+\frac{1}{100}-\frac{1}{103}\right)\)

\(B=\frac{2}{3}\cdot\left(\frac{1}{4}-\frac{1}{103}\right)=\frac{2}{3}\cdot\frac{99}{412}=\frac{33}{206}\)

\(A=\frac{1}{1.4}-\frac{1}{4.7}-\frac{1}{7.10}-...-\frac{1}{2011.2014}\)

\(A=\frac{1}{1.4}-\left(\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{2011.2014}\right)\)

Đặt \(B=\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{2011.2014}\)

\(B=\frac{1}{3}.\left(\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{2011.2014}\right)\)

\(B=\frac{1}{3}.\left(\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{2011}-\frac{1}{2014}\right)\)

\(B=\frac{1}{3}.\left(\frac{1}{4}-\frac{1}{2014}\right)\)

\(B=\frac{1}{3}.\frac{1005}{4028}=\frac{335}{4028}\)

\(A=\frac{1}{4}-\frac{335}{4028}=\frac{168}{1007}\)

A = \(\frac{1}{1.4}-\frac{1}{4.7}-\frac{1}{7.10}-...-\frac{1}{2011.2014}\)

A = 1 + \(\frac{1}{4}\) - \(\frac{1}{4}\) + \(\frac{1}{7}\) - \(\frac{1}{7}\) + \(\frac{1}{10}\) -....- \(\frac{1}{2011}\) + \(\frac{1}{2014}\)

A = 1 + \(\frac{1}{2014}\) = \(\frac{2015}{2014}\)