bài này có tính chất rồi mà 

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

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

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

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

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

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

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

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

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

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

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

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

Ta có : 

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

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

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

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

\(=\)\(2-\frac{1}{17}\)

\(=\)\(\frac{35}{17}\)

Vậy \(\frac{12}{2.5}+\frac{12}{5.8}+\frac{12}{8.11}+...+\frac{12}{65.68}=\frac{35}{17}\)

Chúc bạn học tốt ~ 

= 6/2 - 6/5 + 6/5 - 6/8 + ... + 6/62 - 6/65

= 6/2 - 6/65 = 189/65

6/2 - 6/5 + 6/5 - 6/8 + 6/8 - 6/11 + .... + 6/59 - 6/62 + 6/62 - 6/65

= 6/2 - 6/65

= 189/65

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

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

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

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

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

=1/2-1/5+1/5-1/8+1/8-1/11+1/11-1/14

=1/2-1/14

=7/14-1/14=6/14=3/7

Chịu r

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