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5 tháng 8 2016

\(=\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+........+\frac{2}{x\left(x+1\right)}=\frac{2011}{2013}\)

\(\Rightarrow\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+........+\frac{2}{x.\left(x+1\right)}=\frac{2011}{2013}\)

\(=2.\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+........+\frac{1}{x.\left(x+1\right)}\right)=\frac{2011}{2013}\)

\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+.........+\frac{1}{x}-\frac{1}{x+1}=\frac{2011}{2013}:2\)

\(=\frac{1}{2}-\frac{1}{x+1}=\frac{2011}{4026}\Rightarrow\frac{1}{x+1}=\frac{1}{2}-\frac{2011}{4026}=\frac{1}{2013}\)

\(\Rightarrow x+1=2013\Rightarrow x=2012\)

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

5 tháng 1 2020

Ta có : \(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=\frac{2011}{2013}\)

=> \(2\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2011}{2013}\)

=> \(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}=\frac{2011}{4026}\)

=> \(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2011}{4026}\)

=> \(\frac{1}{2}-\frac{1}{x+1}=\frac{2011}{4016}\Rightarrow\frac{1}{x+1}=\frac{1}{2013}\Rightarrow x+1=2013\Rightarrow x=2012\)

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

30 tháng 9 2016

biết còn hỏi vậy bạn

23 tháng 8 2018

\(\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+...+\dfrac{2}{x\left(x+1\right)}=\dfrac{2011}{2013}\)

\(\Rightarrow2.\left(\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{x\left(x+1\right)}\right)=\dfrac{2011}{2013}\)

\(\Rightarrow\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{x}-\dfrac{1}{x+1}=\dfrac{2011}{4026}\)

\(\Rightarrow\dfrac{1}{2}-\dfrac{1}{x+1}=\dfrac{2011}{4026}\)

\(\Rightarrow\dfrac{1}{x+1}=\dfrac{1}{2013}\)

\(\Rightarrow x+1=2013\)

\(\Rightarrow x=2012\)

Chúc hok dốt!

23 tháng 8 2018

\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x.\left(x+1\right)}=\frac{2011}{2013}\)

\(\Leftrightarrow\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{x.\left(x+1\right)}=\frac{2011}{2013}\)

\(\Rightarrow\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+...+\frac{2}{x\left(x+1\right)}=\frac{2011}{2013}\)

\(\Rightarrow2.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2011}{2013}\)

\(\Rightarrow2.\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2011}{2013}\)

\(\Rightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{2011}{2013}:2\)

\(\Rightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{2011}{4026}\)

\(\Rightarrow\frac{1}{x+1}=\frac{1}{2}-\frac{2011}{4026}\)

\(\Rightarrow\frac{1}{x+1}=\frac{1}{2013}\)

\(\Rightarrow x+1=2013\)

\(\Rightarrow x=2013-1\)

\(\Rightarrow x=2012\)

23 tháng 8 2018

\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x.\left(x+1\right)}=\frac{2011}{2013}\)

\(\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{x.\left(x+1\right)}=\frac{2011}{2013}\) (1/3=2/6;...)

\(\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+...+\frac{2}{x.\left(x+1\right)}=\frac{2011}{2013}\)

\(2.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2011}{2013}\)

\(2.\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2011}{2013}\)

\(1-\frac{2}{x+1}=\frac{2011}{2013}\)

\(\frac{2}{x+1}=\frac{2}{2013}\)

=> x + 1 = 2013

x = 2012

26 tháng 8 2016

\(\frac{2}{6}+\frac{2}{12}+...+\frac{2}{x.\left(x+1\right)}=\frac{2011}{2013}\)

\(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2011}{2013}:2\)

\(\frac{1}{2}-\frac{1}{x+1}=\frac{2011}{4026}\)

\(\frac{\left(x+1-2\right)}{2.\left(x+1\right)}=\frac{2011}{4026}\)