Ta có :
\(1+\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{x\left(x+1\right)}=\frac{2003}{2005}\)
\(\Leftrightarrow\)\(1+2\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2003}{2005}\)
\(\Leftrightarrow\)\(1+2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2003}{2005}\)
\(\Leftrightarrow\)\(1+2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2003}{2005}\)
\(\Leftrightarrow\)\(1+2\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2003}{2005}\)
\(\Leftrightarrow\)\(1+1-\frac{2}{x+1}=\frac{2003}{2005}\)
\(\Leftrightarrow\)\(\frac{2}{x+1}=2-\frac{2003}{2005}\)
\(\Leftrightarrow\)\(\frac{2}{x+1}=\frac{2007}{2005}\)
\(\Leftrightarrow\)\(x+1=2:\frac{2007}{2005}\)
\(\Leftrightarrow\)\(x+1=\frac{4010}{2007}\)
\(\Leftrightarrow\)\(x=\frac{4010}{2007}-1\)
\(\Leftrightarrow\)\(x=\frac{2003}{2007}\)
Vậy \(x=\frac{2003}{2007}\)
Chúc bạn học tốt ~
