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25 tháng 8 2023

Sửa đề:

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

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

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

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

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

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

25 tháng 8 2023

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 = \(\dfrac{11}{17}\)

20 tháng 8 2023

\(\dfrac{3}{2}A=\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+...+\dfrac{3}{94.97}\)

\(\dfrac{3}{2}A=\dfrac{4-1}{1.4}+\dfrac{7-4}{4.7}+\dfrac{10-7}{7.10}+...+\dfrac{97-94}{94.97}\)

\(\dfrac{3}{2}A=\dfrac{4}{1.4}-\dfrac{1}{1.4}+\dfrac{7}{4.7}-\dfrac{4}{4.7}+\dfrac{10}{7.10}-\dfrac{7}{7.10}+...+\dfrac{97}{94.97}-\dfrac{94}{94.97}\)

\(\dfrac{3}{2}A=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{94}-\dfrac{1}{97}\)

\(\dfrac{3}{2}A=1-\dfrac{1}{97}=\dfrac{96}{97}\)

⇒ A = \(\dfrac{96}{97}:\dfrac{3}{2}=\dfrac{64}{97}\)

Câu B cách làm tương tự, thắc mắc gì bạn cứ hỏi nhé.

25 tháng 8 2023

\(A=\dfrac{4}{1\cdot5}+\dfrac{4}{5\cdot9}+...+\dfrac{4}{2001\cdot2005}\)

\(A=1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-...+\dfrac{1}{2001}-\dfrac{1}{2005}\)

\(A=1-\dfrac{1}{2005}=\dfrac{2004}{2005}\)

\(B=\dfrac{3}{10\cdot12}+\dfrac{3}{12\cdot14}+...+\dfrac{3}{998\cdot1000}\)

\(\dfrac{2}{3}B=\dfrac{2}{10\cdot12}+...+\dfrac{2}{998\cdot1000}\)

\(\dfrac{2}{3}B=\dfrac{1}{10}-\dfrac{1}{12}+\dfrac{1}{12}-...+\dfrac{1}{998}-\dfrac{1}{1000}\)

\(\dfrac{2}{3}B=\dfrac{1}{10}-\dfrac{1}{1000}=\dfrac{99}{1000}\)

\(B=\dfrac{99}{1000}:\dfrac{2}{3}=\dfrac{297}{2000}\)

25 tháng 8 2023

\(A=\dfrac{4}{1.5}+\dfrac{4}{5.9}+...+\dfrac{4}{2001.2005}\)

\(\Rightarrow A=4\left(\dfrac{1}{1.5}+\dfrac{1}{5.9}+...+\dfrac{1}{2001.2005}\right)\)

\(\Rightarrow A=4.\dfrac{1}{4}\left(1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+...+\dfrac{1}{2001}-\dfrac{1}{2005}\right)\)

\(\Rightarrow A=1-\dfrac{1}{2005}\)

\(\Rightarrow A=\dfrac{2004}{2005}\)

25 tháng 4 2018

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

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

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

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

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

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

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

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

Vậy A = \(\frac{11}{17}\)

Chúc bạn học tốt!

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

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

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

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

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

24 tháng 7 2021

Thank bạn!

8 tháng 1 2018

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

28 tháng 4 2018

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

28 tháng 4 2018

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

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

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

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

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

9 tháng 8 2015

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

12 tháng 4 2018

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

10 tháng 5 2019

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

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

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

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

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

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

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

~ Hok tốt ~

10 tháng 5 2019

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

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

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

=1/3(3/2*5+3/5*8+...+3/95*98)

=1/3(1/2-1/5+1/5-1/8+...+1/95-1/98)

=1/3*48/98

=1/3*24/49

=8/49