Đáp án:

$\begin{array}{l}
\left| x \right| + \dfrac{1}{{1.2}} + \left| x \right| + \dfrac{1}{{2.3}} + \left| x \right| + \dfrac{1}{{3.4}} + ...\\
 + \left| x \right| + \dfrac{1}{{2019.2020}} = 2020x\\
 \Rightarrow \left( {\left| x \right| + \left| x \right| + ... + \left| x \right|} \right)\\
 + \left( {\dfrac{1}{{1.2}} + \dfrac{1}{{2.3}} + ... + \dfrac{1}{{2019.2020}}} \right) = 2020x\\
 \Rightarrow 2019.\left| x \right| + A = 2020.x\\
Xet:A = \dfrac{1}{{1.2}} + \dfrac{1}{{2.3}} + ... + \dfrac{1}{{2019.2020}}\\
 = \dfrac{{2 - 1}}{{1.2}} + \dfrac{{3 - 2}}{{2.3}} + \dfrac{{4 - 3}}{{3.4}} + ... + \dfrac{{2020 - 2019}}{{2019.2020}}\\
 = 1 - \dfrac{1}{2} + \dfrac{1}{2} - \dfrac{1}{3} + \dfrac{1}{3} - \dfrac{1}{4} + ... + \dfrac{1}{{2019}} - \dfrac{1}{{2020}}\\
 = 1 - \dfrac{1}{{2020}} = \dfrac{{2019}}{{2020}}\\
 \Rightarrow 2019\left| x \right| + \dfrac{{2019}}{{2020}} = 2020x\\
 + Khi:x \ge 0 \Rightarrow \left| x \right| = x\\
 \Rightarrow 2019.x + \dfrac{{2019}}{{2020}} = 2020x\\
 \Rightarrow x = \dfrac{{2019}}{{2020}}\left( {tmdk:x \ge 0} \right)\\
 + Khi:x < 0 \Rightarrow \left| x \right| =  - x\\
 \Rightarrow 2019.\left( { - x} \right) + \dfrac{{2019}}{{2020}} = 2020x\\
 \Rightarrow 4039.x = \dfrac{{2019}}{{2020}}\\
 \Rightarrow x = \dfrac{{2019}}{{2020.4039}} > 0\left( {ktm:x < 0} \right)\\
Vay\,x = \dfrac{{2019}}{{2020}}
\end{array}$