$\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{x.(x+1)}=\dfrac{499}{500}$
$1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{x}-\dfrac{1}{x+1}=\dfrac{499}{500}$
$1-\dfrac{1}{x+1}=\dfrac{499}{500}$
$\dfrac{1}{x+1}=1-\dfrac{499}{500}=\dfrac{1}{500}$
$x+1=500$
$x=500-1=499$