\(A=3\times\left(\frac{3}{1\times4}+\frac{3}{4\times7}+\frac{3}{7\times10}+...+\frac{3}{97\times100}\right)\)

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

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

\(A=3\times\frac{99}{100}\)

\(A=\frac{297}{100}\)

\(A=\frac{3^2}{1.4}+\frac{3^2}{4.7}+\frac{3^2}{7.10}+......+\frac{3^2}{97.100}\)

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

Đặt \(S=\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{97.100}\)

Ta có: \(S=\frac{3}{3}.\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+.....+\frac{3}{97.100}\right)\)

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

\(S=1-\frac{1}{100}=\frac{99}{100}\)

\(\Rightarrow A=3.S=3.\frac{99}{100}=\frac{297}{100}\)

Đặt  \(B=\frac{2}{1\cdot4}+\frac{2}{4\cdot7}+\frac{2}{7\cdot10}+......+\frac{2}{100\cdot103}\)

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

\(B=\frac{2}{3}\cdot\left(1-\frac{1}{103}\right)\)

\(B=\frac{2}{3}\cdot\frac{102}{103}\)

\(\Rightarrow B=\frac{68}{103}\)

Đặt \(A=\frac{2}{1.4}+\frac{2}{4.7}+\frac{2}{7.10}+...+\frac{2}{100.103}\)

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

\(A=\frac{2}{3}\left(1-\frac{1}{103}\right)\)

\(A=\frac{2}{3}\cdot\frac{102}{103}\)

\(A=\frac{68}{103}\)

\(\frac{2}{1\times4}+\frac{2}{4\times7}+\frac{2}{7\times10}+...+\frac{2}{37\times40}\)

\(=\frac{2}{3}\times\left(\frac{3}{1\times4}+\frac{3}{4\times7}+\frac{3}{7\times10}+...+\frac{3}{37\times40}\right)\)

\(=\frac{2}{3}\times\left(\frac{4-1}{1\times4}+\frac{7-4}{4\times7}+\frac{10-7}{7\times10}+...+\frac{40-37}{37\times40}\right)\)

\(=\frac{2}{3}\times\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{37}-\frac{1}{40}\right)\)

\(=\frac{2}{3}\times\left(1-\frac{1}{40}\right)=\frac{13}{20}\)

\(D=\frac{2}{1\cdot4}+\frac{2}{4\cdot7}+\frac{2}{7\cdot10}+...+\frac{2}{97\cdot100}\)

\(D=\frac{2}{3}\left(\frac{3}{1\cdot4}+\frac{3}{4\cdot7}+\frac{3}{7\cdot10}+...+\frac{3}{97\cdot100}\right)\)

\(D=\frac{2}{3}\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{97}-\frac{1}{100}\right)\)

\(D=\frac{2}{3}\left(1-\frac{1}{100}\right)\)

\(D=\frac{2}{3}\cdot\frac{99}{100}=\frac{33}{50}\)

Em cảm ơn chị

Dấu \(.\)là dấu nhân 

Ta có : 

\(E=\frac{2}{1.4}+\frac{2}{4.7}+\frac{2}{7.10}+...+\frac{2}{100.103}\)

\(\Rightarrow E=\frac{2}{3}.\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{2}{100.103}\right)\)

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

\(\Rightarrow E=\frac{2}{3}.\left(1-\frac{1}{103}\right)\)

\(\Rightarrow E=\frac{2}{3}.\frac{102}{103}\)

\(\Rightarrow E=\frac{68}{103}\)

Vậy \(E=\frac{68}{103}\)

~ Ủng hộ nhé 

\(E=\frac{2}{1\cdot4}+\frac{2}{4\cdot7}+...+\frac{2}{100\cdot103}\)

\(E=2\cdot\left(\frac{1}{1\cdot4}+\frac{1}{4\cdot7}+...+\frac{1}{100\cdot103}\right)\)

Gọi tổng trong ngoặc là F

\(\Rightarrow3F=\frac{3}{1\cdot4}+\frac{3}{4\cdot7}+...+\frac{3}{100\cdot103}\)

\(\Rightarrow3F=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{100}-\frac{1}{103}\)

\(\Rightarrow3F=1-\frac{1}{103}=\frac{102}{103}\)

\(\Rightarrow F=\frac{102}{103\cdot3}=\frac{34}{103}\)

\(\Leftrightarrow E=2\cdot\frac{34}{103}=\frac{68}{103}\)

Vậy......

a) \(\dfrac{2}{1\times4}+\dfrac{2}{4\times7}+\dfrac{2}{7\times10}+...+\dfrac{2}{97\times100}\)

\(=2.\left(\dfrac{1}{1\times4}+\dfrac{1}{4\times7}+\dfrac{1}{7\times10}+...+\dfrac{1}{97\times100}\right)\)

\(=2.\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{97}-\dfrac{1}{100}\right)\)

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

\(=2.\dfrac{99}{100}\)

\(=\dfrac{99}{50}\)

_____

b) \(\dfrac{3}{1\times5}+\dfrac{3}{5\times9}+\dfrac{3}{9\times13}+...+\dfrac{3}{97\times101}\)

\(=3.\left(\dfrac{1}{1\times5}+\dfrac{1}{5\times9}+\dfrac{1}{9\times13}+...+\dfrac{1}{97\times101}\right)\)

\(=3.\left(1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{13}+...+\dfrac{1}{97}-\dfrac{1}{101}\right)\)

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

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

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

dấu " . "  là dấu nhân nha bn, nãy viết nhanh quên lớp