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20 tháng 8 2016

1 + 1/3 + 1/6 + 1/10 + ... + 2/x(x + 1) = 4007/2004

2/2 + 2/6 + 2/12 + 2/20 + ... + 2/x(x + 1) = 4007/2004

2 × (1/1×2 + 1/2×3 + 1/3×4 + 1/4×5 + ... + 1/x(x + 1)) = 4007/2004

1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + 1/4 - 1/5 + ... + 1/x - 1/x + 1 = 4007/2004 : 2

1 - 1/x + 1 = 4007/2004 × 1/2

x/x + 1 = 4007/4008

=> x = 4007

1 tháng 3 2019

\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=\frac{2003}{2005}\)

\(\Leftrightarrow\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{x\left(x+1\right)}=\frac{2003}{2005}\)

\(\Leftrightarrow2\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2003}{2005}\)

\(\Leftrightarrow2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2003}{2005}\)

\(\Leftrightarrow2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2003}{2005}\)

\(\Leftrightarrow2\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2003}{2005}\)

\(\Leftrightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{2003}{4010}\)

\(\Leftrightarrow\frac{x+1-2}{2\left(x+1\right)}=\frac{2003}{4010}\)

\(\Leftrightarrow2003.2\left(x+1\right)=4010\left(x-1\right)\)

\(\Leftrightarrow4006x+4006=4010x-4010\)

\(\Leftrightarrow-4x=-8016\)

\(\Leftrightarrow x=2004\)

Vậy x = 2004

1 tháng 3 2019

\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x.\left(x+1\right)}=\frac{2003}{2005}\)

\(\left(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x.\left(x+1\right)}\right).\frac{1}{2}=\frac{2003}{2005}.\frac{1}{2}\)

\(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{2}{x.\left(x+1\right).2}=\frac{2003}{4020}\)

\(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x.\left(x+1\right)}=\frac{2003}{4020}\)

\(\frac{3-2}{2.3}+\frac{4-3}{3.4}+\frac{5-4}{4.5}+...+\frac{\left(x+1\right)-x}{x.\left(x+1\right)}=\frac{2003}{4020}\)

\(\frac{3}{2.3}-\frac{2}{2.3}+\frac{4}{3.4}-\frac{3}{3.4}+...+\frac{x+1}{\left(x+1\right).x}-\frac{x}{\left(x+1\right).x}=\frac{2003}{4020}\)

\(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{\left(x+1\right)}=\frac{2003}{4020}\)

\(\frac{1}{2}-\frac{1}{x+1}=\frac{2003}{4020}\)

\(\frac{1}{x+1}=\frac{1}{2}-\frac{2003}{4020}=\frac{7}{4020}\)

\(\frac{7}{\left(x+1\right).7}=\frac{7}{4020}\)

\(\left(x+1\right).7=4020\)

\(\Rightarrow x=....\)

đặt a=1/3+1/6+1/10+...........+2/n(n+1)

1/2a=1/6+1/12+...........+1/n(n+1)

1/2a=1/2.3+1/3.4+........+1/n(n+1)

1/2a=1/2-1/3+1/3-1/4+.......+1/n-1/n+1

1/2a=1/2-1/n+1

a=(1/2--1/n+1):1/2=2003/2004

1/2-1/n+1=2003/2004.1/2

1/2-1/n+1=2003/4008

1/n+1=1/2-2003/4008

1/n+1=1/4008

suy ra n+1=4008

n=4007

17 tháng 3 2017

n=4007 do

19 tháng 4 2017

\(1+\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=1\frac{2003}{2005}\)

\(\frac{2}{2}+\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{x\left(x+1\right)}=\frac{4008}{2005}\)

\(2.\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{x\left(x+1\right)}\right)=\frac{4008}{2005}\)

\(2.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{4008}{2005}\)

\(=>2.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{4008}{2005}\)

\(2.\left(1-\frac{1}{x+1}\right)=\frac{4008}{2005}\)

=> \(1-\frac{1}{x+1}=\frac{4008}{2005}:2=\frac{2004}{2005}\)

\(\frac{1}{x+1}=1-\frac{2004}{2005}=\frac{1}{2005}\)

=>x+1=2005

=>x=2004

28 tháng 4 2017

1/3 + 1/6 + 1/10 +...+ 2/x(x+1) = 2014/2015

AH
Akai Haruma
Giáo viên
12 tháng 9

Lời giải:

$\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{x(x+1)}=\frac{2004}{2005}$

$2(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x(x+1)})=\frac{2004}{2005}$

$\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x(x+1)}= \frac{1002}{2005}$

$\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x(x+1)}=\frac{1002}{2005}$

$\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+....+\frac{1}{x}-\frac{1}{x+1}=\frac{1002}{2005}$

$\frac{1}{2}-\frac{1}{x+1}=\frac{1002}{2005}$

$\frac{1}{x+1}=\frac{1}{2}-\frac{1002}{2005}=\frac{1}{4010}$

$\Rightarrow x+1=4010$

$\Rightarrow x=4009$

AH
Akai Haruma
Giáo viên
31 tháng 8

Lời giải:

$\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+....+\frac{2}{x(x+1)}=\frac{2004}{2005}$

$\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+....+\frac{2}{x(x+1)}=\frac{2004}{2005}$

$2\left[\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x(x+1)}\right]=\frac{2004}{2005}$
$\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x(x+1)}=\frac{1002}{2005}$

$\frac{3-2}{2.3}+\frac{4-3}{3.4}+\frac{5-4}{4.5}+...+\frac{(x+1)-x}{x(x+1)}=\frac{1002}{2005}$

$\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{x}-\frac{1}{x+1}=\frac{1002}{2005}$

$\frac{1}{2}-\frac{1}{x+1}=\frac{1002}{2005}$

$\frac{1}{x+1}=\frac{1}{2}-\frac{1002}{2005}=\frac{1}{4010}$
$x+1=4010$

$x=4010-1=4009$

 

13 tháng 2 2015

Ta có 1/3+1/6+1/10+....+1/x(x+1)=2004/2005

=>2/6+2/12+2/20+....+2/2x(x+1)=2004/2005

=>2[1/6+1/12+1/20+.......+1/2x(x+1)]=2004/2005

=> 2[1/2.3+1/3.4+1/4.5+.....+1/2x(x+1)] = 2004/2005

=>2[1/2 - 1/3+1/3 -1/4+1/4 - 1/5 +.....+1/2x - 1/(2x+2)] = 2004/2005

=>2[1/2 - 1/(2x+2)] = 2004/2005

=>x/(x+1) = 2004/2005 => x=2004