\(\Leftrightarrow\dfrac{2}{2}+\dfrac{2}{6}+\dfrac{2}{12}+\dfrac{2}{20}+...+\dfrac{2}{x\left(x+1\right)}=1+\dfrac{1999}{1993}\)
\(\Leftrightarrow2\left(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+...+\dfrac{1}{x\left(x+1\right)}\right)=\dfrac{3992}{1993}\)
\(\Leftrightarrow1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{x}-\dfrac{1}{x+1}=\dfrac{1996}{1993}\)
\(\Leftrightarrow1-\dfrac{1}{x+1}=\dfrac{1996}{1993}\)
\(\Leftrightarrow\dfrac{1}{x+1}=\dfrac{-3}{1993}\)
=>x+1=-1993/3
hay x=-1996/3