\(\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{x.\left(x-1\right)}=\frac{2007}{2009}\)

\(2.\left(\frac{1}{2}-\frac{1}{3}\right)+2.\left(\frac{1}{3}-\frac{1}{4}\right)+....+2.\left(\frac{1}{x-1}-\frac{1}{x}\right)=\frac{2007}{2009}\)

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

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

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

\(\frac{1}{x}=\frac{1}{2009}\)

x=2009

22.3+23.4+...+2x.(x−1)=20072009

2.(12−13)+2.(13−14)+....+2.(1x−1−1x)=20072009

2.12−2.1x=20072009

12−1x=20074018

1x=12−20074018

1x=12009

x=2009

kết quả phải là 2008 (cách làm của mình dài dòng lắm)

= 12009 nha mình đúng đấy mình nha 

thanks

\(\frac{1}{2.3}=\frac{1}{2}-\frac{1}{3}\) trong đó 3 = x ; 2 = x - 1 

\(\frac{1}{\left(x-1\right)x}=\frac{1}{x-1}-\frac{1}{x}\)

ĐẶt A = \(\frac{1}{3}+\frac{1}{6}+...+\frac{2}{x\left(x-1\right)}-\frac{2007}{2009}\)

      A = \(\frac{2}{6}+\frac{2}{12}+\frac{2}{30}\cdot\cdot\cdot+\frac{2}{x\left(x-1\right)}-\frac{2007}{2009}\)

      A = \(\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+..+\frac{2}{\left(x-1\right)x}-\frac{2007}{2009}\)

      A = \(\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x-1}-\frac{1}{x}-\frac{2007}{2009}\)

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

\(\frac{2}{6}+\frac{2}{12}+...+\frac{2}{x\left(x-1\right)}=\frac{2007}{2009}\)

\(2\cdot\left(\frac{1}{6}+\frac{1}{12}+..+\frac{1}{x\left(x-1\right)}\right)=\frac{2007}{2009}\)

\(\frac{1}{6}+\frac{1}{12}+..+\frac{1}{x\left(x-1\right)}=\frac{2007}{2009}:2\)

\(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{\left(x-1\right)x}=\frac{2007}{4018}\)

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

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

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

\(\frac{1}{x}=\frac{1}{2009}\)

=> x = 2009