Lời giải:
$\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+....+\frac{2}{x(x+1)}=\frac{2004}{2005}$
$\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+....+\frac{2}{x(x+1)}=\frac{2004}{2005}$
$2\left[\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x(x+1)}\right]=\frac{2004}{2005}$
$\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x(x+1)}=\frac{1002}{2005}$
$\frac{3-2}{2.3}+\frac{4-3}{3.4}+\frac{5-4}{4.5}+...+\frac{(x+1)-x}{x(x+1)}=\frac{1002}{2005}$
$\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{x}-\frac{1}{x+1}=\frac{1002}{2005}$
$\frac{1}{2}-\frac{1}{x+1}=\frac{1002}{2005}$
$\frac{1}{x+1}=\frac{1}{2}-\frac{1002}{2005}=\frac{1}{4010}$
$x+1=4010$
$x=4010-1=4009$
