Tìm x biết:
\(\frac{x}{2013}-\frac{1}{10}-\frac{1}{15}-\frac{1}{21}-.....-\frac{1}{120}=\frac{5}{8}\)
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DT
1
14 tháng 2 2017
\(\Rightarrow\frac{x}{2013}-\left(\frac{2}{20}+\frac{2}{30}+...+\frac{2}{240}\right)=\frac{5}{8}\)
\(\Rightarrow\frac{x}{2013}-2\left(\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{15.16}\right)=\frac{5}{8}\)
\(\Rightarrow\frac{x}{2013}-2\left(\frac{1}{4}-\frac{1}{16}\right)=\frac{5}{8}\)
\(\Rightarrow\frac{x}{2013}-\frac{3}{8}=\frac{5}{8}\)
\(\Rightarrow\frac{x}{2013}=1\)
\(\Rightarrow x=2013\)
NT
4
4 tháng 3 2017
Ta có:
\(\frac{x}{2013}\)-\(\frac{1}{10}\)-\(\frac{1}{15}\)-\(\frac{1}{21}\)-...-\(\frac{1}{120}\)=\(\frac{5}{8}\)
=>\(\frac{x}{2013}\)- (\(\frac{2}{20}\)+\(\frac{2}{30}\)+\(\frac{2}{42}\)+...+\(\frac{2}{240}\)) = \(\frac{5}{8}\)
=>\(\frac{x}{2013}\)- 2.(\(\frac{1}{4.5}\)+\(\frac{1}{5.6}\)+...+\(\frac{1}{15.16}\)) = \(\frac{5}{8}\)
=>\(\frac{x}{2013}\)- 2.(\(\frac{1}{4}\)-\(\frac{1}{10}\)) = \(\frac{5}{8}\)
=>\(\frac{x}{2013}\)- 2.\(\frac{3}{10}\)= \(\frac{5}{8}\)
=>\(\frac{x}{2013}\)= \(\frac{5}{8}\)+\(\frac{6}{10}\)= 1
=> \(x=2013\)
Vậy \(x=2013\)
25 tháng 6 2016
\(\frac{x}{2008}-\frac{1}{10}-\frac{1}{15}-\frac{1}{21}-...-\frac{1}{120}=\frac{5}{8}\)
\(\frac{x}{2008}-\left(\frac{1}{10}+\frac{1}{15}+\frac{1}{21}+...+\frac{1}{120}\right)=\frac{5}{8}\)
\(\frac{x}{2008}-\left(\frac{2}{20}+\frac{2}{30}+\frac{2}{42}+...+\frac{2}{240}\right)=\frac{5}{8}\)
\(\frac{x}{2008}-2.\left(\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+...+\frac{1}{240}\right)=\frac{5}{8}\)
\(\frac{x}{2008}-2.\left(\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{15.16}\right)=\frac{5}{8}\)
\(\frac{x}{2008}-2.\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{15}-\frac{1}{16}\right)=\frac{5}{8}\)
\(\frac{x}{2008}-2.\left(\frac{1}{4}-\frac{1}{16}\right)=\frac{5}{8}\)
\(\frac{x}{2008}-2.\frac{3}{16}=\frac{5}{8}\)
\(\frac{x}{2008}-\frac{3}{8}=\frac{5}{8}\)
\(\frac{x}{2008}=\frac{5}{8}+\frac{3}{8}\)
\(\frac{x}{2008}=1=\frac{2008}{2008}\)
=> x = 2008
Vậy x = 2008
NH
2
LT
22 tháng 9 2019
2008x−101−151−...−1201=85
\frac{x}{2008}-2.\left(\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{15.16}\right)=\frac{5}{8}2008x−2.(4.51+5.61+...+15.161)=85
\frac{x}{2008}-2.\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+....+\frac{1}{15}-\frac{1}{16}\right)=\frac{5}{8}2008x−2.(41−51+51−61+....+151−161)=85
\frac{x}{2008}-2.\left(\frac{1}{4}-\frac{1}{16}\right)=\frac{5}{8}2008x−2.(41−161)=85
\frac{x}{2008}-\frac{3}{8}=\frac{5}{8}2008x−83=85
=> \frac{x}{2008}=\frac{5}{8}+\frac{3}{8}=1=\frac{2008}{2008}2008x=85+83=1=20082008
=> x = 2008
NT
0
VM
12 tháng 11 2019
Ta có:
\(\Rightarrow\frac{x}{2008}=1\)
\(\Rightarrow x=1.2008\)
\(\Rightarrow x=2008\)
Vậy \(x=2008.\)
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