A=\(\frac{7}{10}\)*(\(\frac{10}{3\cdot13}\)+\(\frac{10}{13\cdot23}\)+\(\frac{10}{23\cdot33}\)+\(\frac{10}{43\cdot53}\)+\(\frac{10}{53\cdot63}\))

A=\(\frac{7}{10}\)*(\(\frac{1}{3}\)-\(\frac{1}{13}\)+\(\frac{1}{13}\)-\(\frac{1}{23}\)+\(\frac{1}{23}\)-\(\frac{1}{33}\)+\(\frac{1}{43}\)-\(\frac{1}{53}\)+\(\frac{1}{53}\)-\(\frac{1}{63}\))

A=\(\frac{7}{10}\)*(\(\frac{1}{3}\)-\(\frac{1}{33}\)+\(\frac{1}{43}\)-\(\frac{1}{63}\))

A=\(\frac{7}{10}\)*(\(\frac{10}{33}\)+\(\frac{20}{2709}\))

A=\(\frac{7}{10}\)*\(\frac{9250}{29799}\)

A=\(\frac{925}{4257}\)

trình bày sai

mà cũng giống giống

\(B=\frac{7}{3.13}+\frac{7}{13.23}+...+\frac{7}{53.63}\)

\(B=10.\left(\frac{1}{3.13}+\frac{1}{13.23}+....+\frac{1}{53.63}\right)\)

\(B=10.\left(\frac{1}{3}-\frac{1}{13}+\frac{1}{13}-\frac{1}{23}+...+\frac{1}{53}+\frac{1}{63}\right)\)

\(B=10.\left(\frac{1}{3}-\frac{1}{63}\right)\)

\(B=10.\frac{20}{63}\)

\(B=\frac{200}{63}\)

Cảm ơn bn nha

là khoảng cách : 13 - 3 = 10 thì ta cho 10 là mẫu số của 7, bn hiểu ko???

A \(=\frac{7}{3.13}+\frac{7}{13.23}+...+\frac{7}{53.63}=\frac{7}{10}.\left(\frac{10}{3.13}+\frac{10}{13.23}+...+\frac{10}{53.63}\right)\)

   \(=\frac{7}{10}.\left(\frac{1}{3}-\frac{1}{13}+\frac{1}{13}-\frac{1}{23}+...+\frac{1}{53}-\frac{1}{63}\right)=\frac{7}{10}.\left(\frac{1}{3}-\frac{1}{63}\right)\)

   \(=\frac{7}{10}.\frac{20}{63}=\frac{2}{9}\)

đúng cái nhé

có ai bảo bn sai đâu ?               

\(\frac{7}{3.13}+\frac{7}{13.23}+\frac{7}{23.33}+\frac{7}{33.43}\)

\(=\frac{7}{10}\left(\frac{10}{3.13}+\frac{10}{13.23}+\frac{10}{23.33}+\frac{10}{33.43}\right)\)

\(=\frac{7}{10}\left(\frac{1}{3}-\frac{1}{13}+\frac{1}{13}-\frac{1}{23}+\frac{1}{23}-\frac{1}{33}+\frac{1}{33}-\frac{1}{43}\right)\)

\(=\frac{7}{10}\left(\frac{1}{3}-\frac{1}{43}\right)\)

\(=\frac{7}{10}\left(\frac{43}{129}-\frac{3}{129}\right)\)

\(=\frac{7}{10}.\frac{40}{129}\)

\(=\frac{28}{129}\)

mk làm đúng rồi nha, ko tin bấm thử máy tính

 7/3.13 + 7/13.23 + 7/23.33 + 7/33.43

= 7/10.(1/3-1/13+1/13-1/23+1/23-1/33+1/33-1/43)

= 7/10.(1/3-1/43)

= 7/10 . 14/43

= 49/215

Đặt A ta có :

\(A=\frac{7}{3\times13}+\frac{7}{13\times23}+...+\frac{7}{53\times56}\)

\(A=\frac{7}{10}\times\left(\frac{10}{3\times13}+\frac{10}{13\times23}+...+\frac{10}{53\times56}\right)\)

\(A=\frac{7}{10}\times\left(\frac{1}{3}-\frac{1}{13}+\frac{1}{13}-\frac{1}{23}+...+\frac{1}{53}-\frac{1}{63}\right)\)

\(A=\frac{7}{10}\times\left(\frac{1}{3}-\frac{1}{63}\right)\)

\(A=\frac{7}{10}\times\frac{20}{63}\)

\(\Rightarrow A=\frac{2}{9}\)

A=7/10.(1/3 - 1/13 + 1/13 - 1/23 +.......+ 1/53 - 1/63)

   = 7/10 .( 1/3 - 1/63 )

   =7/10 . 20/63

   = 2/9

A=7/10.(1/3 - 1/13 + 1/13 - 1/23 +.......+ 1/53 - 1/63)

A = 7/10 .( 1/3 - 1/63 )

A   =7/10 . 20/63

A  = 2/9

D

D.2/9

phần a dễ bạn tự làm đi tử thì bạn tính như bình thường còn mẫu thì:7.(\(\frac{1}{3.13}\)+\(\frac{1}{13.23}\)+\(\frac{1}{23.33}\))

\(\frac{7}{10}\).(\(\frac{1}{3}\)-\(\frac{1}{33}\))=\(\frac{7}{33}\)

b)(1+1/3+1/5+..+1/199)-(1/2+1/4+...+1/200)

(1+1/2+1/3+...+1/199+1/200)-(1/2+1/2+1/4+1/4+...+1/200+1/200)

=1+1/2+1/3+...+1/199+1/200-(1+1/2+1/3+...+1/100)

=1/101+1/102+...+1/200

https://olm.vn/hoi-dap/question/60726.html

a)\(\frac{243}{22}\)

b)\(\frac{4}{5}=0,8\)

\(S=\frac{5}{3.13}+\frac{5}{13.23}+.....+\frac{5}{83.93}\)

\(2S=\frac{2.5}{3.13}+\frac{2.5}{13.23}+....+\frac{2.5}{83.93}\)

\(2S=\frac{10}{3.13}+\frac{10}{13.23}+.....+\frac{10}{83.93}\)

\(2S=\frac{1}{3}-\frac{1}{13}+\frac{1}{13}-\frac{1}{23}+....+\frac{1}{83}-\frac{1}{93}\)

\(2S=\frac{1}{3}-\frac{1}{93}=\frac{30}{93}\)

\(S=\frac{30}{93}.\frac{1}{2}=\frac{15}{93}\)

Sửa đề:

\(S=\frac{5}{3.13}+\frac{5}{13.23}+.....+\frac{5}{83.93}\)

\(S=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{13}+\frac{1}{13}-\frac{1}{23}+....+\frac{1}{83}-\frac{1}{93}\right)\)

\(S=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{93}\right)\)

\(S=\frac{1}{2}.\left(\frac{31}{93}-\frac{1}{93}\right)\)

\(S=\frac{1}{2}.\frac{10}{31}\)

\(S=\frac{5}{31}\)

bn lớp 5 ak