(Mẹo: Thông thường thì p/s cuối cùng sẽ là quy luật của tất cả các p/s trong dãy này)
Giải:
Ta có: \(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=\frac{2007}{2009}\)
\(\Leftrightarrow\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{x\left(x+1\right)}=\frac{2007}{2009}\)
\(\Leftrightarrow2\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2007}{2009}\)
\(\Leftrightarrow2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2007}{2009}\)
\(\Leftrightarrow2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2007}{2009}\)
\(\Leftrightarrow2\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2007}{2009}\)
\(\Leftrightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{2007}{2009}:2=\frac{2007}{2009}.\frac{1}{2}=\frac{2007}{4018}\)
\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{2}-\frac{2007}{4018}=\frac{2009-2007}{4018}=\frac{1}{2009}\)
\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{2009}\Rightarrow x+1=2009\)\(\Leftrightarrow x=2009-1=2008\)
Vậy: x=2008
