Sửa đề:

\(A=\dfrac{4}{2.5}+\dfrac{4}{5.8}+\dfrac{4}{8.11}+...+\dfrac{4}{65.68}\)

\(A=4.\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+...+\dfrac{1}{65}-\dfrac{1}{68}\right)\)

\(A=4.\left(\dfrac{1}{2}-\dfrac{1}{68}\right)\)

\(A=4.\left(\dfrac{34}{68}-\dfrac{1}{68}\right)\)

\(A=4.\dfrac{33}{68}\)

\(A=\dfrac{33}{17}\)

A = \(\dfrac{4}{2.5}\) + \(\dfrac{4}{5.8}\)+ \(\dfrac{4}{8.11}\)+...+ \(\dfrac{4}{65.68}\)

A = \(\dfrac{4}{3}\).( \(\dfrac{3}{2.5}\) + \(\dfrac{3}{5.8}\)+ \(\dfrac{3}{8.11}\)+....+ \(\dfrac{3}{65.68}\))

A = \(\dfrac{4}{3}\).(\(\dfrac{1}{2}\) - \(\dfrac{1}{5}\) + \(\dfrac{1}{5}\) - \(\dfrac{1}{8}\) + \(\dfrac{1}{8}\) - \(\dfrac{1}{11}\)+...+ \(\dfrac{1}{65}\)- \(\dfrac{1}{68}\)

A = \(\dfrac{4}{3}\).(\(\dfrac{1}{2}\) - \(\dfrac{1}{68}\))

A = \(\dfrac{4}{3}\). \(\dfrac{33}{68}\)

A = \(\dfrac{11}{17}\)

bạn kiểm tra lại đề nhé.

Ta có: \(A=\dfrac{4}{2\cdot5}+\dfrac{4}{5\cdot8}+...+\dfrac{4}{65\cdot68}\)

\(=\dfrac{4}{3}\cdot\left(\dfrac{3}{2\cdot5}+\dfrac{3}{5\cdot8}+...+\dfrac{3}{65\cdot68}\right)\)

\(=\dfrac{4}{3}\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+...+\dfrac{1}{65}-\dfrac{1}{68}\right)\)

\(=\dfrac{4}{3}\left(\dfrac{1}{2}-\dfrac{1}{68}\right)\)

\(=\dfrac{4}{3}\cdot\dfrac{33}{68}=\dfrac{11}{17}\)

A = \(\dfrac{4}{2.5}\) + \(\dfrac{4}{5.8}\) + \(\dfrac{4}{8.11}\) + ... + \(\dfrac{4}{65.68}\)

7A = \(\dfrac{4.3}{2.5}\) + \(\dfrac{4.3}{5.8}\) + \(\dfrac{4.3}{8.11}\) + ... + \(\dfrac{4.3}{65.68}\)

7A = 4 (\(\dfrac{3}{2.5}\) + \(\dfrac{3}{5.8}\) + \(\dfrac{3}{8.11}\) + ... + \(\dfrac{3}{65.68}\))

7A = 4 (\(\dfrac{1}{2}\) - \(\dfrac{1}{5}\) + \(\dfrac{1}{5}\) - \(\dfrac{1}{8}\) + \(\dfrac{1}{8}\) - \(\dfrac{1}{11}\) + ... + \(\dfrac{1}{65}\) - \(\dfrac{1}{68}\))

7A = 4 (\(\dfrac{1}{2}\) - \(\dfrac{1}{68}\))

7A = 4 . \(\dfrac{33}{68}\) = \(\dfrac{33}{17}\)

A = \(\dfrac{33}{17}\) : 7

=> A = \(\dfrac{33}{119}\)

Ta có: \(A=\dfrac{4}{2.5}+\dfrac{4}{5.8}+\dfrac{4}{8.11}+...+\dfrac{4}{65.68}\)

\(=\dfrac{4}{3}\left(\dfrac{3}{2.5}+\dfrac{3}{5.8}+\dfrac{3}{8.11}+...+\dfrac{3}{65.68}\right)\)

\(=\dfrac{4}{3}\left(\dfrac{5-2}{2.5}+\dfrac{8-5}{5.8}+\dfrac{11-8}{8.11}+...+\dfrac{68-65}{65.68}\right)\)

\(=\dfrac{4}{3}\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+...+\dfrac{1}{65}-\dfrac{1}{68}\right)\)

\(=\dfrac{4}{3}\left(\dfrac{1}{2}-\dfrac{1}{68}\right)=\dfrac{4}{3}.\dfrac{33}{68}=\dfrac{11}{17}\)

ko biết

S = 4/2.5 + 4/5.8 + 4/8.11 + ... + 4/65.48

S = 4/3 . ( 3/2.5 + 3/5.8 + 3/8.11 + ... + 3/65.68 )

S = 4/3 . ( 1/2 - 1/5 + 1/5 - 1/8 + 1/8 - 1/11 + ... + 1/65 - 1/68 )

S = 4/3 . ( 1/2 - 1/68 )

S = 4/3 . 33/68

S = 11/17

\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\)

\(=\frac{11}{17}\)

\(A=\frac{4}{2.5}+\frac{4}{5.8}+\frac{4}{8.11}+...+\frac{4}{65.68}\)

\(A=\frac{4}{3}.\left(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{65.68}\right)\)

\(A=\frac{4}{3}.\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{65}-\frac{1}{68}\right)\)

\(A=\frac{4}{3}.\left[\frac{1}{2}+\left(\frac{1}{5}-\frac{1}{5}\right)+\left(\frac{1}{8}-\frac{1}{8}\right)+...+\left(\frac{1}{65}-\frac{1}{65}\right)-\frac{1}{68}\right]\)

\(A=\frac{4}{3}.\left[\frac{1}{2}-\frac{1}{68}\right]\)

\(A=\frac{4}{3}.\frac{33}{68}\)

\(A=\frac{11}{17}\)

~ Hok tốt ~

\(A=\frac{4}{3}\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+....+\frac{1}{65}-\frac{1}{68}\right)\)

     \(=\frac{4}{3}\left(\frac{1}{2}-\frac{1}{68}\right)\)

       \(=\frac{4}{3}\times\frac{33}{68}=\frac{11}{17}\)

\(A=\frac{4}{2.5}+\frac{4}{5.8}+\frac{4}{8.11}+.........+\frac{4}{65.68}\)

\(A=4\left(\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+.........+\frac{1}{65.68}\right)\)

\(A=\frac{4}{3}\left(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+..........+\frac{3}{65.68}\right)\)

\(A=\frac{4}{3}\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-...........-\frac{1}{65}-\frac{1}{68}\right)\)

\(A=\frac{4}{3}\left(\frac{1}{2}-\frac{1}{68}\right)\)

\(A=\frac{4}{3}\left(\frac{34}{68}-\frac{1}{68}\right)\)

\(A=\frac{4}{3}\left(\frac{33}{68}\right)\)

\(A=\frac{11}{17}\)

Vậy A = \(\frac{11}{17}\)

Chúc bạn học tốt!

a) tìm tất cả các phân số có tử bằng 15 lớn hơn 3/7 và nhỏ hơn 5/8

b) tính tổng S = 4/2.5 + 4/5.8 + 4/8.11 + ... 4/65.68

c) chứng tỏ rằng 16n + 5 / 24n + 7 là phân số tối giản với mọi n thuộc z

Toán lớp 6

  ai tích mình tích lại nh nha

\(A=\frac{4}{2.5}+\frac{4}{5.8}+\frac{4}{8.11}+........+\frac{4}{65.68}\)

\(A=4\left(\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+......+\frac{1}{65.68}\right)\)

\(A=\frac{4}{3}\left(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+..........+\frac{3}{65.68}\right)\)

\(A=\frac{4}{3}\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-.........-\frac{1}{68}\right)\)

\(A=\frac{4}{3}\left(\frac{1}{2}-\frac{1}{68}\right)\)

\(A=\frac{4}{3}\left(\frac{34}{68}-\frac{1}{68}\right)\)

\(A=\frac{4}{3}.\frac{33}{68}\)

\(A=\frac{11}{17}\)

\(\dfrac{12}{2.5}+\dfrac{12}{5.8}+.......+\dfrac{12}{65.68}\)

\(=4\left(\dfrac{3}{2.5}+\dfrac{3}{5.8}+......+\dfrac{3}{65.68}\right)\)

\(=4\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+.......+\dfrac{1}{65}-\dfrac{1}{68}\right)\)

\(=4\left(\dfrac{1}{2}-\dfrac{1}{68}\right)\)

\(=4.\dfrac{33}{68}=\dfrac{33}{17}\)

Dễ quá! Vì mình là một CTV bên Học toán với OnlineMath nên bài này easy!! :")))

\(\dfrac{12}{2.5}+\dfrac{12}{5.8}+\dfrac{12}{8.11}+...+\dfrac{12}{65.68}\)

\(\Leftrightarrow4\left(\dfrac{2}{2.5}+\dfrac{2}{5.8}+\dfrac{2}{8.11}+...+\dfrac{2}{65.68}\right)\)

\(\Leftrightarrow4\left(1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+...+\dfrac{1}{65}-\dfrac{1}{68}\right)\)

\(=4\left(1-\dfrac{1}{68}\right)=4.\dfrac{67}{68}=\dfrac{67}{17}\)

Sorry giải nhầm bài rồi, tính giải một bài khác mà đi ấn nhầm bài này =((( bạn giải theo ctv kia nhé