Đáp án:
\[x = 1992\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
1 + \frac{1}{3} + \frac{1}{6} + \frac{1}{{10}} + .... + \frac{1}{{\frac{{x\left( {x + 1} \right)}}{2}}} = 1 + \frac{{1991}}{{1993}}\\
\Leftrightarrow 2.\left( {\frac{1}{2} + \frac{1}{6} + \frac{1}{{12}} + \frac{1}{{20}} + .... + \frac{1}{{x\left( {x + 1} \right)}}} \right) = 1 + \frac{{1991}}{{1993}}\\
\Leftrightarrow 2.\left( {\frac{1}{{1.2}} + \frac{1}{{2.3}} + \frac{1}{{3.4}} + \frac{1}{{4.5}} + ..... + \frac{1}{{x\left( {x + 1} \right)}}} \right) = 1 + \frac{{1991}}{{1993}}\\
\Leftrightarrow 2.\left( {\frac{{2 - 1}}{{1.2}} + \frac{{3 - 2}}{{2.3}} + \frac{{4 - 3}}{{3.4}} + \frac{{5 - 4}}{{4.5}} + .... + \frac{{\left( {x + 1} \right) - x}}{{x\left( {x + 1} \right)}}} \right) = 1 + \frac{{1991}}{{1993}}\\
\Leftrightarrow 2.\left( {1 - \frac{1}{2} + \frac{1}{2} - \frac{1}{3} + \frac{1}{3} - \frac{1}{4} + \frac{1}{4} - \frac{1}{5} + .... + \frac{1}{x} - \frac{1}{{x + 1}}} \right) = 1 + \frac{{1991}}{{1993}}\\
\Leftrightarrow 2\left( {1 - \frac{1}{{x + 1}}} \right) = 1 + \frac{{1991}}{{1993}}\\
\Leftrightarrow 2 - \frac{2}{{x + 1}} = 1 + \frac{{1991}}{{1993}}\\
\Leftrightarrow \frac{2}{{x + 1}} = 1 - \frac{{1991}}{{1993}}\\
\Leftrightarrow \frac{2}{{x + 1}} = \frac{2}{{1993}}\\
\Leftrightarrow x + 1 = 1993\\
\Leftrightarrow x = 1992
\end{array}\)
