Đề có vấn đề. Bạn coi lại.

1.1/3+1/6+1/10+...+2/x.(x+1)=2007/2009

=>2/6+2/12+2/20+...+2/x.(x+1)=2007/2009

=>1/2-1/3+1/3-1/4+1/4-1/5+...+1/x-1/(x+1)=2007/2009:2

=>1/2-1/(x+1)=2007/4018

=>1/(x+1)=1/2-2007/4018

=>1/x+1=1/2009

=>x+1=2009

=>x=2009-2008

=>x=1

vậy x=1

làm đúng rồi nhưng phần: 

x+1=2009

x=2009-1

x=2008

mà bạn

Đặt A= \(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\times\left(x+1\right)}=\frac{2007}{2009}\)

    \(\Rightarrow\) \(\frac{1}{2}A=\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\times\left(x+1\right)}=\frac{2007}{4018}\)  

\(\Leftrightarrow\frac{1}{2}A=\frac{1}{2\times3}+\frac{1}{3\times4}+\frac{1}{4\times5}+...+\frac{1}{x\times\left(x+1\right)}=\frac{2007}{4018}\)

\(\Leftrightarrow\frac{1}{2}A=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2007}{4018}\)

\(\Leftrightarrow\frac{1}{2}A=\frac{1}{2}-\frac{1}{x+1}=\frac{2007}{4018}\)

\(\Rightarrow A=1-\frac{2}{x+1}=\frac{2007}{2009}\)

\(\Rightarrow\frac{2}{x+1}=1-\frac{2007}{2009}\)

\(\Leftrightarrow\frac{2}{x+1}=\frac{2}{2009}\)

\(\Rightarrow x+1=2009\)

\(\Leftrightarrow x=2009-1\)

\(\Leftrightarrow x=2008\)

Vậy x=2008

\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=\frac{2009}{2011}\)

\(\Rightarrow\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+....+\frac{2}{x\left(x+1\right)}=\frac{2009}{2011}\)

\(\Rightarrow2.\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2009}{2011}\)

\(\Rightarrow2.\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+.....+\frac{1}{x\left(x+1\right)}\right)=\frac{2009}{2011}\)

\(\Rightarrow2.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+....+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2009}{2011}\)

\(\Rightarrow2.\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2009}{2011}\Rightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{2009}{4022}\Rightarrow\frac{1}{x+1}=\frac{1}{2011}\Rightarrow x+1=2011\Rightarrow x=2010\)

Vậy x=2010

\(\frac{2}{6}+\frac{2}{12}+...+\frac{2}{x\left(x+1\right):2}=\frac{2009}{2011}\)

\(2\left(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2009}{2011}\)

\(2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2009}{2011}\)

\(2\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2009}{2011}\)

\(\frac{1}{2}-\frac{1}{x+1}=\frac{2009}{2011}:2\)

\(\frac{1}{2}-\frac{1}{x+1}=\frac{2009}{4022}\)

\(\frac{1}{x+1}=\frac{1}{2}-\frac{2009}{4022}\)

\(\frac{1}{x+1}=\frac{1}{2011}\)

=>x+1=2011

=>x=2010

\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=\frac{2009}{2011}\)

\(\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{x\left(x+1\right)}=\frac{2009}{2011}\)

\(2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2009}{2011}\)

\(2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2009}{2011}\)

\(2\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2009}{2011}\)

\(\frac{1}{2}-\frac{1}{x+1}=\frac{2009}{4022}\)

\(\frac{1}{x+1}=\frac{1}{2011}\)

\(x+1=2011\)

\(x=2010\)

Nhân cả 2 vế với 1/2 ta có:

1/6+ 1/12+ 1/20+... +1/x(x+1)=2009/4022

1/2.3+ 1/3.4+ 1/4.5+... +1/x(x+1)=2009/4022

1/2- 1/3+ 1/3-1/4+ 1/4- 1/5+...+1/x- 1/x+1=2009/4022

1/2- 1/x+1=2009/4022

x+1/2.(x+1)- 2/2.(x+1)=2009/4022

x+1-2/2.(x+1)=2009/4022

x-1/2.(x+1)=2009/4022

=>(x-1). 4022= 2009.2.(x+1)

 4022.x- 4022= 4018.x+ 4018

 4022.x-4018.x= 4018+ 4022

 x.(4022- 4018)= 8040

 x.4                = 8040

 x                   = 8040: 4

 x                   =2010