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3 tháng 8 2015
=> \(\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{x\left(x+1\right)}=\frac{2011}{2013}\)
=> \(\frac{2}{2\times3}+\frac{2}{3\times4}+\frac{2}{4\times5}+...+\frac{2}{x\times\left(x+1\right)}=\frac{2011}{2013}\)
=> \(2\times\left(\frac{1}{2\times3}+\frac{1}{3\times4}+\frac{1}{4\times5}+...+\frac{1}{x\times\left(x+1\right)}\right)=\frac{2011}{2013}\)
=> \(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2011}{2013}:2\)
=> \(\frac{1}{2}-\frac{1}{x+1}=\frac{2011}{4026}\)=> \(\frac{1}{x+1}=\frac{1}{2}-\frac{2011}{4026}=\frac{1}{2013}\)
=> x+1 = 2013 => x = 2012
PN
1
NV
1
1 tháng 6 2018
Cái này lớp 6 :
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+......+\frac{2}{x\left(x+1\right)}=\frac{2011}{2013}\)
\(\Leftrightarrow\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+....+\frac{2}{x\left(x+1\right)}=\frac{2011}{2013}\)
\(\Leftrightarrow\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+....+\frac{2}{x\left(x+1\right)}=\frac{2011}{2013}\)
\(\Leftrightarrow2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+....+\frac{1}{x\left(x+1\right)}\right)=\frac{2011}{2013}\)
\(\Leftrightarrow2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+.....+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2011}{2013}\)
\(\Leftrightarrow2\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2011}{2013}\)
<=> \(\frac{1}{2}-\frac{1}{x+1}=\frac{2011}{4026}\)
\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{2}-\frac{2011}{4026}\)
\(\Leftrightarrow\frac{1}{x+1}=\frac{2}{4026}=\frac{1}{2013}\)
\(\Leftrightarrow x+1=2013\)
=> x = 2012
AK
1 tháng 6 2018
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=\frac{2011}{2013}\)
\(\Rightarrow2.\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2011}{2013}\)
\(\Rightarrow2.\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2011}{2013}\)
\(\Rightarrow2.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2011}{2013}\)
\(\Rightarrow2.\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2011}{2013}\)
\(\Rightarrow1-\frac{2}{x+1}=\frac{2011}{2013}\)
\(\Rightarrow\frac{2}{x+1}=1-\frac{2011}{2013}\)
\(\Rightarrow\frac{2}{x+1}=\frac{2}{2013}\)
\(\Rightarrow x+1=2013\)
\(\Rightarrow x=2013-1\)
\(\Rightarrow x=2012\)
Vậy \(x=2012\)
~ Ủng hộ nhé
XO
14 tháng 8 2019
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{x\times\left(x+1\right):2}=\frac{2011}{2013}\)
\(\Rightarrow\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{x\times\left(x+1\right)}\times\frac{1}{2}=\frac{2011}{2013}\)
\(\Rightarrow2\times\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\times\left(x+1\right)}\right)=\frac{2011}{2013}\)
\(\Rightarrow2\times\left(\frac{1}{2\times3}+\frac{1}{3\times4}+\frac{1}{4\times5}+...+\frac{1}{x\times\left(x+1\right)}\right)=\frac{2011}{2013}\)
\(\Rightarrow2\times\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2011}{2013}\)
\(\Rightarrow2\times\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2011}{2013}\)
\(\Rightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{2011}{2013}:2\)
\(\Rightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{2011}{4026}\)
\(\Rightarrow\frac{1}{x+1}=\frac{1}{2}-\frac{2011}{4016}\)
\(\Rightarrow\frac{1}{x+1}=\frac{1}{2013}\)
\(\Rightarrow x+1=2013\)
\(\Rightarrow x=2012\)
Vậy x = 2012
DP
18 tháng 7 2017
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=\frac{2011}{2013}\)
\(\Leftrightarrow\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+....+\frac{2}{x\left(x+1\right)}=\frac{2011}{2013}\)
\(\Leftrightarrow\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+...+\frac{2}{x\left(x+1\right)}=\frac{2011}{2013}\)
\(\Leftrightarrow2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2011}{2013}\)
\(\Leftrightarrow2\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2011}{2013}\)
\(\Leftrightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{2011}{2013}\div2\)
\(\Leftrightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{2011}{4026}\)
\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{2}-\frac{2011}{4026}\)
\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{2013}\)
\(\Leftrightarrow x=2013-1=2012\)
29 tháng 5 2018
Ta có : 1 x 2 x 3 x ..... x 2012 x 2013 - 1 x 3 x 5 x ..... x 2011 x 2013
= (1 x 3 x 5 x ..... x 2013) x (2 x 4 x 6 x ..... x 2012) - 1 x 3 x 5 x ..... x 2011 x 2013
= (1 x 3 x 5 x ..... x 2011 x 2013) x (2 x 4 x 6 x ..... x 2012 - 1)
\(\dfrac{1}{3}\) + \(\dfrac{1}{6}\) + \(\dfrac{1}{10}\) + ... + \(\dfrac{2}{x:\left(x+1\right)}\) = \(\dfrac{2011}{2013}\)
\(\dfrac{1}{2}\) \(\times\) (\(\dfrac{1}{3}\) + \(\dfrac{1}{6}\) + \(\dfrac{1}{10}\) + ... + \(\dfrac{2}{x}:\left(x+1\right)\) = \(\dfrac{2011}{2013}\) \(\times\) \(\dfrac{1}{2}\)
\(\dfrac{1}{2\times3}\) + \(\dfrac{1}{12}\) + \(\dfrac{1}{20}\) + ... + \(\dfrac{2}{2x\times\left(x+1\right)}\) = \(\dfrac{2011}{2013}\) \(\times\) \(\dfrac{1}{2}\)
\(\dfrac{1}{2\times3}\) + \(\dfrac{1}{3\times4}\) + \(\dfrac{1}{4\times5}\) + ... + \(\dfrac{1}{x\times\left(x+1\right)}\) = \(\dfrac{2011}{2013}\) \(\times\) \(\dfrac{1}{2}\)
\(\dfrac{1}{2}\) - \(\dfrac{1}{3}\) + \(\dfrac{1}{3}\) - \(\dfrac{1}{4}\) + \(\dfrac{1}{4}\) - \(\dfrac{1}{5}\) + ... + \(\dfrac{1}{x}\) - \(\dfrac{1}{x+1}\) = \(\dfrac{2011}{2013}\) \(\times\) \(\dfrac{1}{2}\)
\(\dfrac{1}{2}\) - \(\dfrac{1}{x+1}\) = \(\dfrac{2011}{2013}\) \(\times\) \(\dfrac{1}{2}\)
\(\dfrac{1}{x+1}\) = \(\dfrac{1}{2}\) - \(\dfrac{2011}{2013\times2}\)
\(\dfrac{1}{x+1}\) = \(\dfrac{2013-2011}{2\times2013}\)
\(\dfrac{1}{x+1}\) = \(\dfrac{2}{2\times2013}\)
\(\dfrac{1}{x+1}\) = \(\dfrac{1}{2013}\)
\(x\) + 1 = 2013
\(x\) = 2013 - 1
\(x\) = 2012
Lời giải:
$\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x(x+1)}=\frac{2011}{2013}$
$\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+....+\frac{2}{x(x+1)}=\frac{2011}{2013}$
$\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+....+\frac{2}{x(x+1)}=\frac{2011}{2013}$
$2\left(\frac{3-2}{2.3}+\frac{4-3}{3.4}+\frac{5-4}{4.5}+....+\frac{x+1-x}{x(x+1)}\right)=\frac{2011}{2013}$
$2(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1})=\frac{2011}{2013}$
$2(\frac{1}{2}-\frac{1}{x+1})=\frac{2011}{2013}$
$\frac{1}{2}-\frac{1}{x+1}=\frac{2011}{2013}:2=\frac{2011}{4026}$
$\frac{1}{x+1}=\frac{1}{2}-\frac{2011}{4026}=\frac{1}{2013}$
$x+1=2013$
$x=2013-1$
$x=2012$