Trả lời
$\dfrac{1}{[\dfrac{1}{2}\times{(x+1)}]}$
$=\dfrac{2}{[x(x+1)]}$
$=\dfrac{\dfrac{2}{x-2}}{(x+1)}$.
\Rightarrow Phương trình tương đương: $\dfrac{2}{2}-\dfrac{2}{3}+\dfrac{2}{3}-\dfrac{2}{4}+\dfrac{2}{4}-\dfrac{2}{5}+.....+\dfrac{\dfrac{2}{x-2}}{(x+1)}=\dfrac{1999}{2001}$
\Leftrightarrow $\dfrac{1-2}{(x+1)}=\dfrac{1999}{2001}$
\Leftrightarrow $\dfrac{(x-1)}{(x+1)}=\dfrac{1999}{2001}$
\Rightarrow $x=2000$.