Đáp án:
$\begin{array}{l}
2 + \dfrac{2}{3} + \dfrac{2}{6} + \dfrac{2}{{12}} + ... + \dfrac{2}{{x\left( {x + 1} \right)}}\\
= \left( {2 + \dfrac{2}{3}} \right) + 2.\left( {\dfrac{1}{6} + \dfrac{1}{{12}} + ... + \dfrac{1}{{x\left( {x + 1} \right)}}} \right)\\
= \dfrac{8}{3} + 2.\left( {\dfrac{1}{{2.3}} + \dfrac{1}{{3.4}} + ... + \dfrac{1}{{x\left( {x + 1} \right)}}} \right)\\
= \dfrac{8}{3} + 2.\left( {\dfrac{1}{2} - \dfrac{1}{3} + \dfrac{1}{3} - \dfrac{1}{4} + ... + \dfrac{1}{x} - \dfrac{1}{{x + 1}}} \right)\\
= \dfrac{8}{3} + 2.\left( {\dfrac{1}{2} - \dfrac{1}{{x + 1}}} \right)\\
= \dfrac{8}{3} + 1 - \dfrac{1}{{x + 1}}\\
\Rightarrow \dfrac{8}{3} + 1 - \dfrac{1}{{x + 1}} = 1\dfrac{{1989}}{{1991}}\\
\Rightarrow \dfrac{8}{3} + 1 - \dfrac{1}{{x + 1}} = 1 + \dfrac{{1989}}{{1991}}\\
\Rightarrow \dfrac{8}{3} - \dfrac{1}{{x + 1}} = \dfrac{{1989}}{{1991}}\\
\Rightarrow \dfrac{1}{{x + 1}} = \dfrac{8}{3} - \dfrac{{1989}}{{1991}}\\
\Rightarrow \dfrac{1}{{x + 1}} = \dfrac{{9961}}{{5973}}\\
\Rightarrow x + 1 = \dfrac{{5973}}{{9961}}\\
\Rightarrow x = \dfrac{{5973}}{{9961}} - 1\\
\Rightarrow x = \dfrac{{ - 3988}}{{9961}}
\end{array}$
