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 

Bài này giống toán lớp 6 hơn

m = 3/(1x4) + 3/(4x7) +  ... + 3/(19x22)

    = (4-1)/(1x4) + (7-4)/(4x7) +  ... + (22-19)/(19x22)

    = 4/(1x4) - 1/(1x4) + 7/(4x7) - 4/(4x7) + ... +  22/(19x22) - 19/(19x22)

    = 1 - 1/4 + 1/4 - 1/7 + ... + 1/19 - 1/22

    = 1-1/22

    = 21/22

A=\(\frac{3}{1.4}+\frac{3}{4.7}+...+\frac{3}{2.\left(x+3\right)}\)

=> A=\(\frac{3}{1}-\frac{3}{4}+\frac{3}{4}+...+\frac{3}{2.x}-\frac{3}{2.\left(x+3\right)}\) 

=> A =\(\frac{3}{1}-\frac{3}{2.\left(x+3\right)}\)

tinh nhanh 1/1x4 + 1/4x7 +1/7x10 +...+ 1/91x94 

Ta có :

1/1.4+1/4.7+...+1/91.94

=1/3.(1/1-1/4+...+1/91-1/94)

=1/3.(1/1-1/94)

=1/3.93/94

=31/94

1/1.4+1/4.7+1/7.10+...+1/91.94

=1/3.(3/1.4+3/4.7+3/7.10+...+3/91.94)

=1/3.(1-1/4+1/4-1/7+1/7-1/10+...+1/91-1/94)

=1/3.(1-1-94)

=1/3.(93/94)

=31/94

Đặt A= 1/1*4+1/4*7+1/7*10+....+1/91*94

3A= 3/1*4+3/4*7+3/7*10+....+3/91*94

3A=1/1-1/4+1/4-1/7+1/7-1/10+............+1/91-1/94

3A=1-1/94=93/94=>A=93/94*1/3=31/94

=31/94 k mình nha bạn