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

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

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

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

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

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=2/3(1-1/4+1/4-1/7+1/7-1/10+...+1/97-1/100)

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

A=2/3.99/100

A=33/50

mình k pit co dung k nua nghe

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

=2/3(3/1.4+3/4.7+3/7.10+...+3/97.100)

=2/3(1-1/4+1/4-1/7+1/7-1/10+...+1/97-1/100)

=2/3(1-1/100)=33/50

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

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

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

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

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

33/50

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

\(B=\frac{2}{1.4}+\frac{2}{4.7}+...+\frac{2}{97.100}=\frac{1}{3}\left(\frac{2}{1}-\frac{2}{4}+\frac{2}{4}-...-\frac{2}{100}\right)\)

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

Bạn ơi tớ hỏi Nguyễn Thiều Công Thành:

Vì sao lại = 2/3 . ( 3/1.4 + 3/4.7+ 3/7/10 + ... + 3/97.100 )  

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

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

=2.(1-1/100)

=2.99/100=99/50

B = 2 (1/1.4 + 1/4.7 + 1/7.10 + ... + 1/97.100 )

B = 2/3 ( 3/1.4 + 3/4.7 + 3/7.10 + ... + 3/97.100 )

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

B = 2/3 ( 1/1 - 1/100 ) = 2/3 . 99/100 = 33/50

Bạn lưu ý ở bước thứ 3 với công thức này d/a.b = 1/a - 1/b với d = a-b. BẠn cứ dùng công thức này mà ko cần giải thích vì công thức này khá phổ biến. Nếu phải giải thích thì bạn cứ dùng công thức này để giải thích.

Ờ bước thứ hai mình làm như vậy vì để đưa về công thức mà mình nói.

Chúc bạn học tốt!

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

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

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

\(=2.\frac{99}{100}\) 

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

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

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

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

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

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

Study well ! >_<

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

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

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

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

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

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

có phải là 99/100 đúng không

mình cần gấp lắm có ai giúp giupf mình với!

Mình ko chắc lắm, nếu sai thì xin lỗi nhiều

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

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

\(A=2.\left(\frac{1.3}{1.4.3}+\frac{1.3}{4.7.3}+\frac{1.3}{7.10.3}+...+\frac{1.3}{97.100.3}\right)\)

\(A=2.\frac{1}{3}.\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{97.100}\right)\)

\(A=2.\frac{1}{3}.\left(\frac{4-1}{1.4}+\frac{7-4}{4.7}+\frac{10-7}{7.10}+...+\frac{100-97}{97.100}\right)\)

\(A=\frac{2}{3}.\left(\frac{4}{1.4}-\frac{1}{1.4}+\frac{7}{4.7}-\frac{4}{4.7}+\frac{10}{7.10}-...-\frac{97}{97.100}\right)\)

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

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

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

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