\(A=\frac{2}{3}\left(\frac{3}{1.4}+\frac{3}{4.7}+...+\frac{3}{97.100}\right)\)

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

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

\(A=\frac{2}{3}.\frac{99}{100}\)

\(A=\frac{33}{50}\)

\(\frac{2}{1.4}+\frac{2}{4.7}+\frac{2}{7.10}+...+\frac{2}{97.100}\)

\(=2\left(\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{97.100}\right)\)

\(=\frac{2}{3}\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{97.100}\right)\)

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

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

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

\(=\frac{33}{50}\)

ủa ko phải câu đầu là như thế này hả bạn  \(A=\frac{2}{3}\left(\frac{2}{1.4}+\frac{2}{4.7}+...+\frac{2}{97.100}\right)\)

B =\(\frac{2}{1.4}+\frac{2}{4.7}+\frac{2}{7.10}+...\frac{2}{97.100}\)

=2.(\(\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{97.100}\))

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

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

3B=2.(1-\(\frac{1}{100}\))

3B=2.\(\frac{99}{100}\)=\(\frac{99}{50}\)

=>B=\(\frac{99}{50}:3\)=\(\frac{33}{50}\)

Tick mik nha

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

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

\(=\frac{2}{3}\left(1-\frac{1}{100}\right)=\frac{33}{50}\)

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

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

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

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

\(A=\frac{2}{3}.\frac{99}{100}\)

\(\Rightarrow A=\frac{33}{50}\)

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

\(\Rightarrow\frac{3}{2}A=\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{100}\)

\(\Rightarrow\frac{3}{2}A=\frac{1}{1}-\frac{1}{100}=\frac{99}{100}\)

\(\Rightarrow A=\frac{99}{100}:\frac{3}{2}=\frac{33}{50}\)

B=2(1/1.4 +1/4.7+....+1/97.100)

3B= 2(3/1.4+3/4.7+...+3/97.100)

3B=2(1-1/4+1/4-1/7+...+1/97-1/100)

3B= 2(1-1/100)

3B= 2.99/100

3B= 99/50

B=33/50.

B=2/1.4+2/4.7+2/7.10+...+2/97.100

B=2/3.3/1.4+2/3.3/4.7+2/3.3/7.10+...+2/3.3/97.100

B=2/3(1-1/4+1/4-1/7+1/7-1/10+...+1/97-1/100) (dùng phương pháp khử)

B=2/3(1-1/100)

B=2/3.99/100

B=33/50

B=\(\frac{2}{1.4}+\frac{2}{4.7}+\frac{2}{7.10}+....+\frac{2}{97.100}\)

B=\(\frac{2}{1}-\frac{2}{4}+\frac{2}{4}-\frac{2}{7}+...+\frac{2}{97}-\frac{2}{100}\)

B=\(\frac{2}{1}-\frac{2}{100}=\frac{200}{100}-\frac{2}{100}\)

B=\(\frac{198}{100}=\frac{46}{25}\)

a) \(\frac{5}{1.4}+\frac{5}{4.7}+\frac{5}{7.10}+.....+\frac{5}{27.30}\)

\(=\frac{5}{3}\left(\frac{1}{1.4}+\frac{1}{4.7}+........+\frac{1}{27.30}\right)\)

\(=\frac{5}{3}\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+.....+\frac{1}{27}-\frac{1}{30}\right)\)

\(=\frac{5}{3}\left(1-\frac{1}{30}\right)\)

\(=\frac{5}{3}.\frac{29}{30}=\frac{29}{36}\)

Đặt \(A=\frac{12}{3\cdot5}+\frac{12}{5\cdot7}+\frac{12}{7\cdot9}+....+\frac{12}{97\cdot99}\)

\(2A=\frac{12}{3}-\frac{12}{5}+\frac{12}{5}-\frac{12}{7}+...+\frac{12}{97}-\frac{12}{99}\)

\(2A=\frac{12}{3}-\frac{12}{99}\)

\(A=\frac{128}{33}\cdot\frac{1}{2}=\frac{64}{33}\)

1.5/1-5/4+5/4-5/7+5/7-5/9 +....+5/27-5/30

=5/1-5/30

=145/30=29/6.

anh ơi ,toán này hồi em học lớp 4 còn biết thế mà anh ko biết, gợi ý nha:toán này thuộc dạng sai phân

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

\(\frac{3}{2}A=1-\frac{1}{100}\)

\(\frac{3}{2}A=\frac{99}{100}\)

\(A=\frac{33}{50}\)

k minh nha

bài này dễ thế mà không giải được hả bạn

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

\(A=\frac{2}{3}.\left(\frac{3}{1.4}+\frac{3}{4.7}+...+\frac{3}{97.100}\right)\)

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

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

\(A=\frac{2}{3}.\frac{99}{100}=\frac{33}{50}\)

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

A = \(\frac{2}{3}.\left(1-\frac{1}{100}\right)\)= \(\frac{2}{3}.\frac{99}{100}\)= \(\frac{33}{50}\)
 

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

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

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

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

A = \(\frac{2}{3}\cdot\frac{99}{100}\)

A = \(\frac{33}{50}\)

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

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

\(=2.\frac{99}{100}\) 

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

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

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

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

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

=> \(A=\frac{2}{3}.\frac{99}{100}=\frac{33}{50}\)

Study well ! >_<

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

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

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

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

\(A=\frac{2}{3}.\frac{99}{100}\)

\(A=\frac{33}{50}\)