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23 tháng 11 2017

\(\dfrac{1}{1.2}\) + \(\dfrac{1}{2.3}\) + \(\dfrac{1}{3.4}\) +...+ \(\dfrac{x}{x.(x+1)}\) = \(\dfrac{44}{45}\)

\(\Rightarrow\) 1 - \(\dfrac{1}{2}\) +\(\dfrac{1}{2}\) - \(\dfrac{1}{3}\) +...+ \(\dfrac{1}{x}\) - \(\dfrac{1}{x+1}\) = \(\dfrac{44}{45}\)

\(\Rightarrow\) 1 - \(\dfrac{1}{x+1}\) = \(\dfrac{44}{45}\)

\(\Rightarrow\) \(\dfrac{x}{x+1}\) = \(\dfrac{44}{45}\) 

\(\Rightarrow\) \(x=44\)

23 tháng 11 2017

<=> 1/2-1/3+1/3-1/4+...+1/x-1/x+1 = 44/45

<=> 1/2-1/x+1 = 44/45

<=> 1/x+1 = 1/2 - 44/45 = -43/90

=> x+1 = -90/43

=> x = -133/43

k mk nha

12 tháng 6 2018

\(\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{8.9.10}\right).x=\frac{23}{45}\)

\(\frac{1}{2}.\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{8.9.10}\right).x=\frac{23}{45}\)

\(\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{8.9}-\frac{1}{9.10}\right).x=\frac{23}{45}\)

\(\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{9.10}\right).x=\frac{23}{45}\)

\(\frac{1}{2}.\frac{22}{45}.x=\frac{23}{45}\)

         \(\frac{11}{45}.x=\frac{23}{45}\)

                  \(x=\frac{23}{45}:\frac{11}{45}\)

                 \(x=\frac{23}{11}\)

12 tháng 6 2018

Gọi A=(1/1.2.3+ 1/2.3.4 +...+ 1/8.9.10) .x=23/45

    2A=3-1/1.2.3+ 4–2/2.3.4+ 5–4/3.4.5+ ... + 10–8/8.9.10

    2A=1/2 —1/2.3+ 1/2.3 — 1/3.4+ 1/3.4– 1/4.5 +...+1/8.9–1/9.10=1/2–1/9.10=44/90

     A=44/90 : 2=22/90

     x=23/45:A= 23/45 : 22/90=23/11= 2 1/1( hỗn số)

    

2 tháng 11 2018

\(\left(\frac{1}{1.2}+\frac{1}{2.3}+......+\frac{1}{9.10}\right)\left(x-1\right)+\frac{1}{10}x=x-\frac{9}{10}\)

\(\Rightarrow\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+......+\frac{1}{9}-\frac{1}{10}\right)\left(x-1\right)+\frac{1}{10}x=x-\frac{9}{10}\)

\(\Rightarrow\left(1-\frac{1}{10}\right)\left(x-1\right)+\frac{1}{10}x=x-\frac{9}{10}\)

\(\Rightarrow\frac{9}{10}.\left(x-1\right)+\frac{1}{10}x=x-\frac{9}{10}\)

\(\Rightarrow\frac{9}{10}x-\frac{9}{10}+\frac{1}{10}x=x-\frac{9}{10}\)

\(\Rightarrow\left(\frac{9}{10}x+\frac{1}{10}x\right)-\frac{9}{10}=x-\frac{9}{10}\)

\(\Rightarrow x-\frac{9}{10}=x-\frac{9}{10}\)

\(\Rightarrow x\inℝ\)

Vậy \(x\inℝ\)

11 tháng 2 2017

Trước hết ta thực hiện biểu thức trong ngoặc:

\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{8.9.10}\)

\(=\frac{1}{2}\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{8.9.10}\right)\)

\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{8.9}-\frac{1}{9.10}\right)\)

\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{9.10}\right)\)

\(=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{90}\right)\)

\(=\frac{1}{2}.\frac{22}{45}\) \(=\frac{11}{45}\)

\(\Rightarrow\frac{11}{45}\) \(.x=\frac{22}{45}\)

\(\Rightarrow x=\frac{22}{45}:\frac{11}{45}\)

\(\Rightarrow x=2\)

29 tháng 1 2016

\(\Leftrightarrow\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+..+\frac{2}{8.9.10}\right).x=\frac{44}{45}\)

\(\Leftrightarrow\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+..+\frac{1}{8.9}-\frac{1}{9.10}\right).x=\frac{44}{45}\)

\(\Leftrightarrow\left(\frac{1}{1.2}-\frac{1}{9.10}\right).x=\frac{44}{45}\Leftrightarrow\frac{22}{45}.x=\frac{44}{5}\Leftrightarrow x=2\)

vậy x=2

29 tháng 1 2016

x=2 nhé

olm đang duyệt

7 tháng 2 2017

\(A=\frac{5}{2.1}+\frac{4}{1.11}+\frac{3}{11.14}+\frac{1}{14.15}+\frac{13}{15.28}\)

\(\frac{A}{7}=\frac{5}{2.7}+\frac{4}{7.11}+\frac{3}{11.14}+\frac{1}{14.15}+\frac{13}{15.28}\)

\(\frac{A}{7}=\frac{7-2}{2.7}+\frac{11-7}{7.11}+\frac{14-11}{11.4}+\frac{15-14}{14.15}+\frac{28-15}{15.28}\)

\(\frac{A}{7}=\frac{1}{2}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{15}+\frac{1}{15}-\frac{1}{28}=\frac{1}{2}-\frac{1}{28}=\frac{13}{28}\)

\(A=7.\frac{13}{28}\)

\(A=\frac{13}{4}\)

18 tháng 11 2019

Ta có

\(\frac{1}{n\left(n+1\right)}=\frac{1}{n}-\frac{1}{n+1}\)   và \(\frac{1}{n\left(n+1\right)\left(n+2\right)}=\frac{1}{n}-\frac{1}{n+1}-\frac{1}{n+2}\)  nên

\(A=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{n\left(n+1\right)}+...+\frac{1}{2008\cdot2009}=1-\frac{1}{2009}=\frac{2008}{2009}\)

\(2B=\frac{2}{1\cdot2\cdot3}+\frac{2}{2\cdot3\cdot4}+...+\frac{2}{n\left(n+1\right)\left(n+2\right)}+...+\frac{2}{2008\cdot2009\cdot2010}\)

\(=\frac{1}{1\cdot2}-\frac{1}{2009\cdot2010}=\frac{201944}{2009\cdot2010}\)

\(\Rightarrow B=\frac{1}{2}\cdot\frac{201944}{2009\cdot2010}=\frac{1009522}{2009\cdot2010}\)

Do đó \(\frac{B}{A}=\frac{1009522}{2009\cdot2010}:\frac{2008}{2009}=\frac{1009522\cdot2009}{2008\cdot2009\cdot2010}=\frac{5047611}{2018040}\)