\(3B=\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{100.103}.\)

\(3B=\frac{4-1}{1.4}+\frac{7-4}{4.7}+\frac{10-7}{7.10}+...+\frac{103-100}{100.103}\)

\(3B=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{100}-\frac{1}{103}=1-\frac{1}{103}=\frac{102}{103}\)

\(B=\frac{102}{3.103}=\frac{34}{103}\)

B=\(\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+......+\frac{1}{100.103}\)

B=\(\frac{1}{3}.\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+....+\frac{3}{100.103}\right)\)

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

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

B=\(\frac{1}{3}.\frac{102}{103}\)

B=\(\frac{34}{103}\)

\(\dfrac{3}{2}A=\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+...+\dfrac{3}{94.97}\)

\(\dfrac{3}{2}A=\dfrac{4-1}{1.4}+\dfrac{7-4}{4.7}+\dfrac{10-7}{7.10}+...+\dfrac{97-94}{94.97}\)

\(\dfrac{3}{2}A=\dfrac{4}{1.4}-\dfrac{1}{1.4}+\dfrac{7}{4.7}-\dfrac{4}{4.7}+\dfrac{10}{7.10}-\dfrac{7}{7.10}+...+\dfrac{97}{94.97}-\dfrac{94}{94.97}\)

\(\dfrac{3}{2}A=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{94}-\dfrac{1}{97}\)

\(\dfrac{3}{2}A=1-\dfrac{1}{97}=\dfrac{96}{97}\)

⇒ A = \(\dfrac{96}{97}:\dfrac{3}{2}=\dfrac{64}{97}\)

Câu B cách làm tương tự, thắc mắc gì bạn cứ hỏi nhé.

Giúp mình đi

Đặt 2/3 ra ngoài  trong ngoặc còn :

1-1/4+1/4-1/7+...-1/97=96/97

Lấy 2/3 nhân với 96/97 sẽ ra đáp án nhé

`S_1 = 5/(1.4) + 5/(4.7) +...+ 5/(97.100)`

`S_1 = 5 (1/(1.4) + 1/(4.7) +...+ 1/(97.100))`

`S_1 = 5/3 (3/(1.4) + 3/(4.7) +...+ 3/(97.100))`

`S_1 = 5/3 (1 - 1/4 + 1/4 - 1/7 + ...+ 1/97 - 1/100)`

`S_1 = 5/3 (1 - 1/100)`

`S_1 = 5/3 . 99/100`

`S_1 = 33/20`

5/1.4 + 5/4.7 + 5/7.10 + ... + 5/97.100

= 5/3 . (3/1.4 + 3/4.7 + 3/7.10 + ... + 3/97.100)

= 5/3 . (1 - 1/4 + 1/4 - 1/7 + 1/7 - 1/10 + ... + 1/97 - 1/100)

= 5/3 . ( 1 - 1/100)

= 5/3 . 99/100

= 33/20

Làm từng phần nha bạn

\(\frac{3}{1\cdot4}+\frac{3}{4\cdot7}+...+\frac{3}{298\cdot301}+x=\frac{299}{301}\)

Đặt \(A+x=\frac{299}{301}\)

\(A=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{298}-\frac{1}{301}\)

\(A=1-\frac{1}{301}\)

\(A=\frac{300}{301}\)

=> \(\frac{300}{301}+x=\frac{299}{301}\)

\(x=\frac{299-300}{301}\)

\(x=-\frac{1}{301}\)

\(A=5\cdot\left(\frac{1}{1\cdot4}+\frac{1}{4\cdot7}+...+\frac{1}{301\cdot304}\right)\)

\(\frac{3A}{5}=\frac{3}{1\cdot4}+\frac{3}{4\cdot7}+...+\frac{3}{301\cdot304}\)

\(\frac{3}{5}\cdot A=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{301}-\frac{1}{304}\)

\(\frac{3}{5}\cdot A=1-\frac{1}{304}\)

\(\frac{3}{5}\cdot A=\frac{303}{304}\)

\(A=\frac{505}{304}\)

Có : 3/5 A = 3/1.4 + 3/4.7 + 3/7.10 + ..... + 3/307.310

= 1 - 1/4 + 1/4 - 1/7 + 1/7 - 1/10 + ........ + 1/307 - 1/310

= 1 - 1/310

= 309/310

=> A = 309/310 : 3/5 = 103/62

Tk mk nha

        \(\frac{5}{1.4}+\frac{5}{4.7}+\frac{5}{7.10}+...+\frac{5}{307.310}\)

\(=\frac{5}{3}\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{307.310}\right)\)

\(=\frac{5}{3}\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{307}-\frac{1}{310}\right)\)

\(=\frac{5}{3}\left(1-\frac{1}{310}\right)\)

\(=\frac{5}{3}.\frac{309}{310}=\frac{103}{62}\)

A=(5/1-5/4+5/4-5/7+...+5/307-5/310):3

A=(5/1-5/310):3

A=103/62

không thích trả lời đấy

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

mik chỉ hỏi thui mà, mik có bắt bạn giúp âu

họ hỏi ko trả lời thì hoi mặc chửi ngt :)