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17
2 tháng 4 2015
\(\Rightarrow\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+....+\frac{2}{x\cdot\left(x+1\right)}=\frac{2013}{2015}\)
\(\Rightarrow2.\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+....+\frac{1}{x\cdot\left(x+1\right)}\right)=\frac{2013}{2015}\)
\(\Rightarrow2.\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+....+\frac{1}{x\cdot\left(x+1\right)}\right)=\frac{2013}{2015}\)
\(\Rightarrow2.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+....+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2013}{2015}\)
\(\Rightarrow\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2013}{2015}:2\)
\(\Rightarrow-\frac{1}{x+1}=\frac{2013}{4030}-\frac{1}{2}\)
\(\Rightarrow-\frac{1}{x+1}=-\frac{1}{2015}\Rightarrow x+1=2015\Rightarrow x=2014\)
NT
2
17 tháng 9 2016
1/2x3/2+1/3x4/2+1/4x5/2+1/5x6/2+.......+2/Xx(X+1)=2011/2013
2/2x3+2/3x4+2/4x5+2/5x6+.....+2/Xx(X+1)=2011/2013
2x(1/2x3+1/3x4+1/4x5+1/5x6+....+1/Xx(x+1)=2011/2013
1/2x3+1/3x4+1/4x5+1/5x6+....+1/Xx(X+1)=2011/4026
1/2-1/3+1/3-1/4+1/4-1/5+1/5-1/6+.....+ 1/x-1/x+1=2011/4026
1/2-1/x+1=2011/4026
1/x+1=1/2-2011/4026
1/x+1=1/2013
Suy ra x=2012
13 tháng 8 2016
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=\frac{2011}{2013}\)
=> \(\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{x\left(x+1\right)}=\frac{2011}{2013}\)
=> \(2.\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2011}{2013}\)
=> \(2.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2011}{2013}\)
=> \(2.\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2011}{2013}\)
=> \(2.\frac{1}{2}-2.\frac{1}{x+1}=\frac{2011}{2013}\)
=> \(1-\frac{2}{x+1}=\frac{2011}{2013}\)
=> \(\frac{2}{x+1}=1-\frac{2011}{2013}=\frac{2}{2013}\)
=> x + 1 = 2013
=> x = 2013 - 1 = 2012
XO
5 tháng 1 2020
Ta có : \(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=\frac{2011}{2013}\)
=> \(2\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2011}{2013}\)
=> \(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}=\frac{2011}{4026}\)
=> \(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2011}{4026}\)
=> \(\frac{1}{2}-\frac{1}{x+1}=\frac{2011}{4016}\Rightarrow\frac{1}{x+1}=\frac{1}{2013}\Rightarrow x+1=2013\Rightarrow x=2012\)