ta thấy

1.4=1(2+2)=1.2+1.2=1.2+2

2.5=2(3+2)=2.3+2.2=2.3+4

......................................

100.103=100(101+2)=100.101+100.2=100.101+200

B=1.2+2+2.3+4+3.4+6+...........................+100.101+200

đặt các phép tính nhân là C còn đặt các số tự nhiên là D

tính D trước khoảng cách các số hạng là 2 

co so số hang là :(200-2):2+1=100 số hạng

D= (200+2).100:2=10100

tính C 

ta thấy

1.2=1.2.3/3

2.3=2.3.4/3

................

100.101=100.101.102/3

triệt tiêu các phân số ta có

100.101.102/3-0=343400

vậy B=C+D=343400+10100=353500 

Ta có: 5/1x4 + 5/4x7 + ... + 5/100x103

= 5/3 x (1/1 - 1/4 + 1/4 - 1/7 +...+1/100 - 1/103)

= 5/3 x (1 - 1/103)

= 5/3 x 102/103

= 170/103

170/103

câu hỏi là gì vậy bạn

tính hay sao í

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......

E = 2/1×4 + 2/4×7 + 2/7×10 + ... + 2/100×103

E = 2( 1/1×4 + 1/4×7 + 1/7×10 + ... + 1/100×103)

3E = 2( 3/1×4 + 3/4×7 + 3/7×10 + ... + 3/100×103 )

3E = 2(1 - 1/4 + 1/4 - 1/7 + 1/7 - 1/10 + ...+ 1/100-1/103)

3E = 2( 1 - 1/103 )

3E = 2 × 102/103

3E = 204/103

=> E = 204/103 : 3

     E = 68/103

Hok tốt

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

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

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

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

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

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

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

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

\(=\frac{2}{3}\cdot\frac{102}{103}=\frac{68}{103}\)

Ta có: 5/1x4 + 5/4x7 + ... + 5/100x103 

B = 5/3 x (1/1 - 1/4 + 1/4 - 1/7 +...+1/100 - 1/103)

B = 5/3 x (1 - 1/103)

B = 5/3 x 102/103

B = 170/103 

Giờ anh đang bận hồi nữa anh giúp cho nha

\(F=\dfrac{1}{x}\left(\dfrac{1}{1.4}+\dfrac{1}{4.7}+...+\dfrac{1}{97.100}+\dfrac{1}{100.103}\right)\)

\(3F=\dfrac{1}{x}\left(\dfrac{3}{1.4}+\dfrac{3}{4.7}+...+\dfrac{3}{97.100}+\dfrac{3}{100.103}\right)\)

\(F=\dfrac{\dfrac{1}{x}\left(\dfrac{1}{3}-\dfrac{1}{103}\right)}{3}=\dfrac{\dfrac{1}{x}.\dfrac{100}{309}}{3}=\dfrac{\dfrac{100x}{309}}{3}=\dfrac{100x}{927}\)

\(F=\dfrac{1}{3}\cdot\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{100}-\dfrac{1}{103}\right)\)

\(=\dfrac{1}{3}\left(1-\dfrac{1}{103}\right)=\dfrac{1}{3}.\dfrac{102}{103}=\dfrac{204}{309}\)

bài toán lớp mấy vậy?

\(A=\dfrac{1}{1\times4}+\dfrac{1}{4\times7}+\dfrac{1}{7\times10}+...+\dfrac{1}{100\times103}\)

\(=\dfrac{1}{3}\left(\dfrac{3}{1\times4}+\dfrac{3}{4\times7}+...+\dfrac{3}{100\times103}\right)\)

\(=\dfrac{1}{3}\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{100}-\dfrac{1}{103}\right)\)

\(=\dfrac{1}{3}\left(1-\dfrac{1}{103}\right)=\dfrac{1}{3}.\dfrac{102}{103}=\dfrac{34}{103}\)