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22 tháng 7 2019

=> D = 1/2-1/5+1/5-1/8+....+ 1/62-1/65

=> D= 1/2-1/65

=> D=63/130

  VẬY D=63/130

22 tháng 7 2019

\(D=\frac{3}{2\cdot5}+\frac{3}{5\cdot8}+\cdot\cdot\cdot+\frac{3}{62\cdot65}\)

\(\Rightarrow D=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\cdot\cdot\cdot+\frac{1}{62}-\frac{1}{65}\)

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

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

18 tháng 7 2017

Đặt \(A=\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+\frac{3}{14.17}\)

\(A=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{17}\)

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

\(A=\frac{15}{34}\)

18 tháng 7 2017

\(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{17}\)\(\frac{1}{2}-\frac{1}{17}\)=\(\frac{15}{34}\)

25 tháng 4 2019

Ta có: \(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{14.17}\)

\(=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{14}-\frac{1}{17}\)

\(=\frac{1}{2}-\frac{1}{17}=\frac{15}{34}\)\(< \frac{17}{34}=\frac{1}{2}\)

\(\Rightarrow\frac{3}{2.5}+\frac{3}{5.8}+...+\frac{3}{14.17}< \frac{1}{2}\)

Vậy:..........................................(đpcm)

25 tháng 4 2019

xét vế trái

ta có:đề\(=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{14}-\frac{1}{17}\) 

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

vậy vế trái bé hơn \(\frac{1}{2}\)

P/S:  \(< < \)là luôn luôn bé hơn nha

k mình nha bạn

Thiengl2015#

25 tháng 4 2019

Ta có :

\(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{14.17}\)

\(=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{14}-\frac{1}{17}\)

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

Mà \(\frac{1}{2}-\frac{1}{17}< \frac{1}{2}\)

Nên \(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{14.17}< \frac{1}{2}\left(đpcm\right)\)

6 tháng 3 2020

\(\frac{3^2}{2.5}+\frac{3^2}{5.8}+\frac{3^2}{8.11}+\frac{3^2}{11.14}+\frac{3^2}{14.17}\)

=\(3\left(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+\frac{3}{14.17}\right)\)

\(3\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{17}\right)\)

\(3\left(\frac{1}{2}-\frac{1}{17}\right)\)

=\(\frac{45}{34}\)

\(3\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{17}\right)\)

=3(3/2.5+3/5.8+3/8.11+3/11.14+3/14.17)

=3(1/2-1/5+1/5-1/8+1/8-1/11+1/11-1/14+1/14-1/17)

=3(1/2-1/17)

=45/34

cô Nhung ơi k đúng cho con đi cô pls

26 tháng 4 2017

\(\frac{3}{2.5}\)+\(\frac{3}{5.8}\)+\(\frac{3}{8.11}\)+\(\frac{3}{11.14}\)+\(\frac{3}{14.17}\)

=\(\frac{1}{2}\)-\(\frac{1}{5}\)+\(\frac{1}{5}\)-\(\frac{1}{8}\)+......+\(\frac{1}{14}\)-\(\frac{1}{17}\)

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

=\(\frac{15}{34}\)

26 tháng 4 2017

bn cho mk đi là câu tl sẽ hiện ra

6 tháng 4 2017

Sai đề => Sửa: \(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{17.20}\)

\(\Rightarrow\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{17}-\frac{1}{20}\)

\(\Rightarrow\frac{1}{2}-\frac{1}{20}\)

\(\Rightarrow\frac{9}{20}\)

6 tháng 4 2017

nói j chứ trả lời sao đc

27 tháng 8 2017

a) = 1-1/2+1/2-1/3+1/3-1/4

    = 1-1/4=3/4

b)=1-1/2+1/2-1/3+1/3-1/4+...+1/2016-1/2017+1/2017-1/2018

   =1-1/2018=2017/2018

c)=1/2-1/5+1/5-1/8+1/8-1/11+1/2009-1/2012+1/2012-1/2015

   = 1/2-1/2015=2015/4030-2/4030=2013/4030

27 tháng 8 2017

a) \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}=1-\frac{1}{4}=\frac{3}{4}\)

b) \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2017.2018}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2017-2018}\)

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

\(=\frac{2017}{2018}\)

c) \(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{2012.2015}\)

\(=3\left(\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{2012.2015}\right)\)

\(\Leftrightarrow\frac{3}{2}\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{2012}-\frac{1}{2015}\right)\)

\(=\frac{3}{2}\left(\frac{1}{2}-\frac{1}{2015}\right)\)

\(=\frac{3}{2}.\frac{2013}{4030}\)

\(=\frac{6039}{8060}\)

\(P=...\)

\(=\frac{1}{99}-\frac{1}{99}+\frac{1}{98}-\frac{1}{98}+\frac{1}{97}-...-\frac{1}{2}+1\)

\(=\frac{1}{99}-1=\frac{-98}{99}\)

\(M=...\)

\(=\frac{2}{2}+\frac{1}{2}+\frac{4}{4}+\frac{1}{4}+...+\frac{64}{64}+\frac{1}{64}-7\)

\(=1+1+1+1+1+1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+\frac{1}{2^5}+\frac{1}{2^6}-7\)

\(=\frac{1+2+2^2+2^3+2^4+2^5}{2^6}-1\)

\(=\frac{2^6-1}{2^6}-1=1-\frac{1}{2^6}-1=-\frac{1}{2^6}\)