Đáp án:
$x=2002$
Giải thích các bước giải:
Ta có:
$\begin{array}{l}
\dfrac{1}{3} + \dfrac{1}{6} + \dfrac{1}{{10}} + ... + \dfrac{1}{{x\left( {x + 1} \right):2}} = \dfrac{{2001}}{{2003}}\\
\Leftrightarrow \dfrac{1}{{2.3:2}} + \dfrac{1}{{3.4:2}} + \dfrac{1}{{4.5:2}} + ... + \dfrac{1}{{x\left( {x + 1} \right):2}} = \dfrac{{2001}}{{2003}}\\
\Leftrightarrow \dfrac{2}{{2.3}} + \dfrac{2}{{3.4}} + \dfrac{2}{{4.5}} + ... + \dfrac{2}{{x\left( {x + 1} \right)}} = \dfrac{{2001}}{{2003}}\\
\Leftrightarrow 2\left( {\dfrac{1}{{2.3}} + \dfrac{1}{{3.4}} + \dfrac{1}{{4.5}} + ... + \dfrac{1}{{x\left( {x + 1} \right)}}} \right) = \dfrac{{2001}}{{2003}}\\
\Leftrightarrow \dfrac{1}{{2.3}} + \dfrac{1}{{3.4}} + \dfrac{1}{{4.5}} + ... + \dfrac{1}{{x\left( {x + 1} \right)}} = \dfrac{{2001}}{{4006}}\\
\Leftrightarrow \dfrac{1}{2} - \dfrac{1}{3} + \dfrac{1}{3} - \dfrac{1}{4} + \dfrac{1}{4} - \dfrac{1}{5} + ... + \dfrac{1}{x} - \dfrac{1}{{x + 1}} = \dfrac{{2001}}{{4006}}\\
\Leftrightarrow \dfrac{1}{2} - \dfrac{1}{{x + 1}} = \dfrac{{2001}}{{4006}}\\
\Leftrightarrow \dfrac{1}{{x + 1}} = \dfrac{1}{{2003}}\\
\Leftrightarrow x + 1 = 2003\\
\Leftrightarrow x = 2002
\end{array}$
Vậy $x=2002$
