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

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

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

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

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

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

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

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

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

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

=> x = 2009 

a. nhân cả hai vế của đẳng thức với 1/ 10 ta có

x/10 - (2/11.13 +2/13.15+...+2/53.55)=3/11 . 1/10

x/10 - (1/11-1/13+1/13-1/15 +...+1/53-1/55) =3/110

x/10 - (1/11 - 1/55) =3/110

x/10 -4/55 = 3/110

x/10=3/110 + 4/55

x. 1/10 =1/10

x= 1/10 : 1/10 =1

b) bạn nhân cả hai vế của đẳng thức với 1/2 rồi làm tương tự

a. nhân cả hai vế của đẳng thức với \(\frac{1}{10}\). Ta có:

\(\frac{x}{10}-\left(\frac{2}{11.13}+\frac{2}{13.15}+...\frac{2}{53.55}\right)=\frac{3}{11}.\frac{1}{10}\)

\(\frac{x}{10}-\left(\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}+...\frac{1}{53}-\frac{1}{55}\right)=\frac{3}{110}\)

\(\frac{x}{10}-\left(\frac{1}{11}-\frac{1}{55}\right)=\frac{3}{110}\)

\(\frac{x}{10}-\frac{-4}{55}=\frac{3}{110}\)

\(\frac{x}{10}=\frac{3}{110}+\frac{4}{55}\)

\(x.\frac{1}{10}=\frac{1}{10}\)

\(x=\frac{1}{10}:\frac{1}{10}=1\)

b. cũng thế bạn nhân hai vế của đẳng thức với \(\frac{1}{2}\) rồi làm tương tự.

Ta có :

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

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

\(\Leftrightarrow\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}=\frac{2007}{4018}\)

\(\Leftrightarrow\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}=\frac{2007}{4018}\)

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

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

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

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

\(\Rightarrow x+1=2009\)

\(\Rightarrow x=2009-1\)

\(\Rightarrow x=2008\)

X = 2008

\(\Rightarrow\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{x\left(x+1\right)}=\frac{2007}{2009}\)

\(\Rightarrow\)\(2.\)(\(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}=\frac{2007}{2009}\)

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

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

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

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

\(\Rightarrow\frac{-1}{x+1}=\frac{-1}{4018}\Rightarrow x+1=4018\Rightarrow x=4017\)

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

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

=> \(\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{x\left(x+1\right)}=\frac{2007}{2009}\)

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

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

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

=> \(\frac{2}{x+1}=\frac{2}{2009}\)   =>  x + 1 = 2009 => x = 2008

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

\(\Rightarrow\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{x\left(x+1\right)}=\frac{2007}{2009}\)

\(\Rightarrow2.\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2007}{2009}\)

\(\Rightarrow2.\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2007}{2009}\)

\(\Rightarrow2.\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2007}{2009}\)

\(\Rightarrow1-\frac{2}{x+1}=\frac{2007}{2009}\)

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

\(\Rightarrow2009=x+1\)

\(\Rightarrow x=2008\)