tính
\(\frac{2}{1.4}\) + \(\frac{2}{4.7}\) + \(\frac{2}{7.10}\) + \(\frac{2}{28.31}\)
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
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
\(A=\frac{2}{1.4}+\frac{2}{4.7}+\frac{2}{7.10}+...+\frac{2}{97.100}\)
\(A=\frac{2}{3}.\left(\frac{3}{1.4}+\frac{3}{4.7}+...+\frac{3}{97.100}\right)\)
\(A=\frac{2}{3}.\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{100}\right)\)
\(A=\frac{2}{3}.\left(1-\frac{1}{100}\right)\)
\(A=\frac{2}{3}.\frac{99}{100}=\frac{33}{50}\)
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 = \(\frac{2}{3}.\left(1-\frac{1}{100}\right)\)= \(\frac{2}{3}.\frac{99}{100}\)= \(\frac{33}{50}\)
\(A=2.\left(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{100}\right)\)
\(=2.\left(\frac{1}{1}-\frac{1}{100}\right)\)
\(=2.\frac{99}{100}\)
\(=\frac{99}{50}\)
\(A=\frac{2}{1.4}+\frac{2}{4.7}+\frac{2}{7.10}+...+\frac{2}{97.100}\)
=> \(A=\frac{2}{3}\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{97.100}\right)\)
=> \(A=\frac{2}{3}\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{97}-\frac{1}{100}\right)\)
=> \(A=\frac{2}{3}\left(1-\frac{1}{100}\right)\)
=> \(A=\frac{2}{3}.\frac{99}{100}=\frac{33}{50}\)
Study well ! >_<
C = 2/1.4 + 2/4.7 + 2/7.10 + .... + 2/601.604
C = 2/3 . ( 3/1.4 + 3/4.7 + 3/7.10 + ... + 3/601.604 )
C = 2/3 . ( 1 - 1/4 + 1/4 - 1/7 + 1/7 - 1/10 + ... + 1/601 - 1/604 )
C = 2/3 . ( 1 - 1/604 )
C = 2/3 . 603/604
C = 201/302
\(C=\frac{2}{1.4}+\frac{2}{4.7}+\frac{2}{7.10}+.....+\frac{2}{601.604}=\frac{2}{3}\cdot\left(\frac{3}{1.4}+\frac{3}{4.7}+...+\frac{3}{601.604}\right)=\frac{2}{3}\cdot\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{601}-\frac{1}{604}\right)\)=\(\frac{2}{3}\cdot\left(1-\frac{1}{604}\right)=\frac{2}{3}\cdot\frac{603}{604}=\frac{201}{302}\)
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
\(B=\frac{2}{1.4}+\frac{2}{4.7}+\frac{2}{7.10}+...+\frac{2}{97.100}=\frac{2}{3}\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{97.100}\right)\)
\(=\frac{2}{3}\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{97}-\frac{1}{100}\right)=\frac{2}{3}\left(1-\frac{1}{100}\right)\)
\(=\frac{2}{3}.\frac{99}{100}=\frac{33}{50}\)
\(B=\frac{2}{1.4}+\frac{2}{4.7}+...+\frac{2}{97.100}=\frac{1}{3}\left(\frac{2}{1}-\frac{2}{4}+\frac{2}{4}-...-\frac{2}{100}\right)\)
\(B=\frac{1}{3}.\left(2-\frac{2}{100}\right)=\frac{1}{3}.\frac{99}{50}==\frac{33}{50}\)
=\(\frac{1}{3}\times\left(\frac{2}{1}-\frac{2}{4}+\frac{2}{4}-\frac{2}{8}+...+\frac{2}{28}-\frac{2}{31}\right)\)
=\(\frac{1}{3}\times\left(\frac{2}{1}-\frac{2}{31}\right)=\frac{20}{31}\)
Bấm đúng cho tui, đi mà. CHÚC BẠN HỌC GIỎI
bài giải đó là sai giả như vầy nè
\(=\frac{1}{3}\cdot2\cdot\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{8}+...+\frac{1}{28}-\frac{1}{31}\right)\)
\(=\frac{2}{3}\left(1-\frac{1}{31}\right)\)
=\(\frac{2}{3}\cdot\frac{30}{31}=\frac{20}{31}\)