[tex] \frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} +...+ \frac{1}{x(x+1)} = \frac{2010}{2011}\\
\frac{1}{1}- \frac{1}{2} + \frac{1}{2} - \frac{1}{3}+...+ \frac{1}{x} - \frac{1}{x+1} =\frac{2010}{2011}\\
1- \frac{1}{x+1}=\frac{2010}{2011}\\
\frac{1}{x+1}=1-\frac{2010}{2011}\\
\frac{1}{x+1}= \frac{1}{2011} \\
x+1=2011\\
x=2010[/tex]