Ta có :
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=\frac{2003}{2004}\)
\(\Leftrightarrow\)\(\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{x\left(x+1\right)}=\frac{2003}{2004}\)
\(\Leftrightarrow\)\(2\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2003}{2004}\)
\(\Leftrightarrow\)\(2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2003}{2004}\)
\(\Leftrightarrow\)\(2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2003}{2004}\)
\(\Leftrightarrow\)\(2\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2003}{2004}\)
\(\Leftrightarrow\)\(\frac{1}{2}-\frac{1}{x+1}=\frac{2003}{2004}:2\)
\(\Leftrightarrow\)\(\frac{1}{2}-\frac{1}{x+1}=\frac{2003}{4008}\)
\(\Leftrightarrow\)\(\frac{1}{x+1}=\frac{1}{2}-\frac{2003}{4008}\)
\(\Leftrightarrow\)\(\frac{1}{x+1}=\frac{1}{4008}\)
\(\Leftrightarrow\)\(x+1=4008\)
\(\Leftrightarrow\)\(x=4007\)
Vậy \(x=4007\)
Chúc bạn học tốt ~
