Let \(A=1+\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+...+\dfrac{1}{x\left(x+1\right):2}=1\dfrac{1991}{1993}\)

\(A=\dfrac{1}{2}\cdot\left(1+\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+...+\dfrac{1}{x\cdot\left(x+1\right):2}\right)=\dfrac{3984}{3986}\)

\(A=\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+...+\dfrac{1}{x\cdot\left(x+1\right)}=\dfrac{3984}{3986}\)

\(A=\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{x}-\dfrac{1}{x+1}=\dfrac{3984}{3986}\)

\(A=\dfrac{1}{1}-\dfrac{1}{x+1}=\dfrac{3984}{3986}\Rightarrow x=1992\)

So x = 1992.