\(=\dfrac{4}{3}\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+...+\dfrac{1}{302}-\dfrac{1}{305}\right)=\dfrac{4}{3}\cdot\dfrac{303}{610}=\dfrac{202}{305}\)

\(\dfrac{4}{2.5}+\dfrac{4}{5.8}+\dfrac{4}{8.11}+...+\dfrac{4}{302.305}\)

\(=4\left(\dfrac{1}{2.5}+\dfrac{1}{5.8}+\dfrac{1}{8.11}+...+\dfrac{1}{302.305}\right)\)

\(=\dfrac{4}{3}\left(\dfrac{3}{2.5}+\dfrac{3}{5.8}+\dfrac{3}{8.11}+...+\dfrac{3}{302.305}\right)\)

\(=\dfrac{4}{3}\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+...+\dfrac{1}{302}-\dfrac{1}{305}\right)\)

\(=\dfrac{4}{3}\left(\dfrac{1}{2}-\dfrac{1}{305}\right)\)

\(=\dfrac{4}{3}.\dfrac{303}{610}\\ =\dfrac{202}{305}\)

\(=\frac{3}{4}\cdot\left(\frac{4}{2.5}+\frac{4}{5.8}+...+\frac{4}{62.65}\right)\)

\(=\frac{3}{2.5}+\frac{3}{5.8}+...+\frac{3}{62.65}\)

\(=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{62}-\frac{1}{65}\)

\(=\frac{1}{2}-\frac{1}{65}\)

\(=\frac{63}{130}\)

Đặt A=4/2.5+4/5.8+4/8.11+...+4/62.65.Ta có A=4.(1/2.5+1/5.8+1/8.11+...1/62.65)=4/3.(3/2.5+3/5.8+3/8.11+...+3/62.65)                                         =4/3.(1/2-1/5+1/5-1/8+1/8-1/11+...+3/62-3/65)=4/3.(1/2-1/65)=4/3.63/130=42/56                                                                                            Vậy A=42/56

M = 4/2.5 + 4/5.8 + 4/8.11 + 4/11.14 + 4/14.17 + 4/17.20

M= 4/3 . (1/2-1/5+1/5-1/8+1/8-1/11+1/11-1/14+1/14-1/17+1/17-1/20)

M= 4/3 . (1/2 - 1/20)

M= 4/3 . (10/20 - 1/20)

M= 4/3 . 9/20

M= 3/5

k nha

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

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

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

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

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

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.

\(A=\dfrac{4}{2\cdot5}+\dfrac{4}{5\cdot8}+\dfrac{4}{8\cdot11}+...+\dfrac{4}{65\cdot68}\\ =\dfrac{4}{3}\cdot\dfrac{3}{2\cdot5}+\dfrac{4}{3}\cdot\dfrac{3}{5\cdot8}+\dfrac{4}{3}\cdot\dfrac{3}{8\cdot11}+...+\dfrac{4}{3}\cdot\dfrac{3}{65\cdot68}\\ =\dfrac{4}{3}\left(\dfrac{3}{2\cdot5}+\dfrac{3}{5\cdot8}+\dfrac{3}{8\cdot11}+...+\dfrac{3}{65\cdot68}\right)\\ =\dfrac{4}{3}\cdot\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+...+\dfrac{1}{65}-\dfrac{1}{68}\right)\\ =\dfrac{4}{3}\cdot\left(\dfrac{1}{2}-\dfrac{1}{68}\right)\\ =\dfrac{4}{3}\cdot\dfrac{33}{68}\\ =\dfrac{11}{17}\)

3/4 . M = 1/2.5 + 1/5.8 + 1/8.11 + 1/11.14 + 1/14.17 + 1/17.20

3/4 . M = 1/2 - 1/20

3/4 .M = 9/10

M = 9/10 : 3/4

M = 3/5

 Vậy M = 3/5

Nhớ tk m nha

\(A=\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{2015.2018}\)

\(=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{2015}-\frac{1}{2018}\)

\(=\frac{1}{2}-\frac{1}{2018}=\frac{504}{1009}\)

Vậy \(A=\frac{504}{1009}.\)

\(B=\frac{4}{2.6}+\frac{4}{6.10}+\frac{4}{10.14}+...+\frac{4}{102.106}\)

\(=\frac{1}{2}-\frac{1}{6}+\frac{1}{6}-\frac{1}{10}+\frac{1}{10}-\frac{1}{14}+...+\frac{1}{102}-\frac{1}{106}\)

\(=\frac{1}{2}-\frac{1}{106}=\frac{26}{53}\)

Vậy \(B=\frac{26}{53}.\)

Bài làm:

a) \(A=\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{2015.2018}\)

\(A=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{2015}-\frac{1}{2018}\)

\(A=\frac{1}{2}-\frac{1}{2018}\)

\(A=\frac{504}{1009}\)

b) \(B=\frac{4}{2.6}+\frac{4}{6.10}+\frac{4}{10.14}+...+\frac{4}{102.106}\)

\(B=\frac{1}{2}-\frac{1}{6}+\frac{1}{6}-\frac{1}{10}+\frac{1}{10}-\frac{1}{14}+...+\frac{1}{102}-\frac{1}{106}\)

\(B=\frac{1}{2}-\frac{1}{106}\)

\(B=\frac{26}{53}\)

\(A=\frac{3}{2.5}+\frac{3}{5.8}+...+\frac{3}{2015.2018}=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{2015}-\frac{1}{2018}\)

\(=\frac{1}{2}-\frac{1}{2018}=\frac{504}{1009}\)

\(B=\frac{4}{2.6}+\frac{4}{6.10}+...+\frac{4}{102.106}=\frac{1}{2}-\frac{1}{6}+\frac{1}{6}-\frac{1}{10}+...+\frac{1}{102}-\frac{1}{106}=\frac{1}{2}-\frac{1}{106}\)

\(=\frac{26}{53}\)