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

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

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

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

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

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

\(\frac{12}{2.5}+\frac{12}{5.8}+\frac{12}{8.11}+...+\frac{12}{29.32}\)

\(=4.\left(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{29.32}\right)\)

\(=4.\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{29}-\frac{1}{32}\right)\)

\(=4.\left(\frac{1}{2}-\frac{1}{32}\right)\)

\(=4.\frac{15}{32}\)

\(=\frac{15}{8}\)

_Chúc bạn học tốt_

\(\frac{12}{2.5}+\frac{12}{5.8}+\frac{12}{8.11}+....+\frac{12}{29.32}\)

\(=4\left(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{29.32}\right)\)

\(=4\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{29}-\frac{1}{32}\right)\)

\(=4\left(\frac{1}{2}-\frac{1}{32}\right)\)

\(=4.\frac{15}{32}=\frac{15}{8}\)

Đặt A= 12/2*5+12/5*8+12/8*11+....+12/29*32

       A= 4*(3/2*5+3/5*8+3/8*11+...+3/29*32)

       A = 4*(1/2-1/5+1/5-1/8+1/8-1/11+....+1/29-1/32)

       A= 4*(1/2-1/32)

       A= 4*15/32

       A= 15/8

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

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

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

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

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

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

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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.

9D=2.5.9+5.8.9+8.11.9+11.14.9+...+62.65.9+65.68.9

9D=2.5.9+5.8.(11-2)+8.11.(14-5)+11.14.(17-8)+...+62.65.(68-59)+65.68.(71-62)

9D=2.5.9+5.8.11-2.5.8+8.11.14-5.8.11+11.14.17-8.11.14+...+62.65.68-59.62.65+65.68.71-62.65.67

Rút gọn các phần giống nhau còn

9D=2.5.9-2.5.8+65.68.71

9D=10+65.68.71

9D=10+313820

9D=313830

D=313830:9=34870

Vậy D=34870

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

ủng hộ mình nha

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

\(\Rightarrow A=\frac{1}{3}\left(\frac{1}{2}-\frac{1}{68}\right)=\frac{1}{2}\left(\frac{34}{68}-\frac{1}{68}\right)=\frac{1}{2}.\frac{33}{68}=\frac{33}{136}\)