Chưa chắc là đề sai!!!!!!!!!!!!!!
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=\frac{2001}{2003}\)
\(\Leftrightarrow\frac{1}{6}+\frac{1}{12}+...+\frac{1}{x\left(x+1\right)}=\frac{2001}{2003}.\frac{1}{2}=\frac{2001}{4006}\)
\(\Leftrightarrow\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...\frac{1}{x}-\frac{1}{x+1}=\frac{2001}{4006}\)
\(\Leftrightarrow\)\(\frac{1}{2}-\frac{1}{x+1}=\frac{2001}{4006}\)
\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{2}-\frac{2001}{4006}\)
\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{2003}\)
\(x+1=2003\)
\(x=2002\)
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=\frac{1999}{2001}\)
\(\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{x\left(x+1\right)}=\frac{1999}{2001}\)
\(2\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{1999}{2001}\)
\(2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{1999}{2001}\)
\(2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x-1}\right)=\frac{1999}{2001}\)
\(2\left(\frac{1}{2}-\frac{1}{x-1}\right)=\frac{1999}{2001}\)
\(\frac{1}{2}-\frac{1}{x-1}=\frac{1999}{2001}:2\)
\(\frac{1}{2}-\frac{1}{x-1}=\frac{1999}{4002}\)
\(\frac{1}{x-1}=\frac{1}{2}-\frac{1999}{4002}\)
\(\frac{1}{x-1}=\frac{1}{2001}\)
\(\Rightarrow x+1=\frac{1.2001}{1}=2001\)
\(\Rightarrow x=2001-1\)
\(\Rightarrow x=2000\)
Vậy \(x=2000\)
