Đáp án:
$x=4007$
Giải thích các bước giải:
$\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+...+\dfrac{2}{x(x+1)}=\dfrac{2003}{2004}\\
\Leftrightarrow \dfrac{1}{2}.\left (\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+...+\dfrac{2}{x(x+1)} \right )=\dfrac{2003}{2004}.\dfrac{1}{2}\\
\Leftrightarrow \dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+...+\dfrac{1}{x(x+1)}=\dfrac{2003}{2004}.\dfrac{1}{2}\\
\Leftrightarrow \dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{x}-\dfrac{1}{x+1}=\dfrac{2003}{4008}\\
\Leftrightarrow \dfrac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\dfrac{1}{x+1}=\dfrac{2003}{4008}\\ \Leftrightarrow \dfrac{1}{2}-\dfrac{1}{x+1}=\dfrac{2003}{4008}\\
\Leftrightarrow \dfrac{1}{x+1}=\dfrac{1}{2}-\dfrac{2003}{4008}\\
\Leftrightarrow \dfrac{1}{x+1}=\dfrac{2004}{4008}-\dfrac{2003}{4008}\\
\Leftrightarrow \dfrac{1}{x+1}=\dfrac{1}{4008}\\
\Leftrightarrow x+1=4008\\
\Leftrightarrow x=4007$
