Đáp án:

\(\left[ \begin{array}{l}
x = \dfrac{{4041}}{2}\\
x = \dfrac{{4033}}{2}
\end{array} \right.\)

Giải thích các bước giải:

\(\begin{array}{l}
\dfrac{{{{\left( {2018 - x} \right)}^2} + \left( {2018 - x} \right)\left( {x - 2019} \right) + {{\left( {x - 2019} \right)}^2}}}{{{{\left( {2018 - x} \right)}^2} - \left( {2018 - x} \right)\left( {x - 2019} \right) + {{\left( {x - 2019} \right)}^2}}} = \dfrac{{19}}{{49}}\\
 \to \dfrac{{{{\left( {2018 - x} \right)}^2} + 2\left( {2018 - x} \right)\left( {x - 2019} \right) + {{\left( {x - 2019} \right)}^2} - \left( {2018 - x} \right)\left( {x - 2019} \right)}}{{{{\left( {2018 - x} \right)}^2} - 2\left( {2018 - x} \right)\left( {x - 2019} \right) + {{\left( {x - 2019} \right)}^2} + \left( {2018 - x} \right)\left( {x - 2019} \right)}} = \dfrac{{19}}{{49}}\\
 \to \dfrac{{{{\left( {2018 - x + x - 2019} \right)}^2} - \left( {2018 - x} \right)\left( {x - 2019} \right)}}{{{{\left( {2018 - x - x + 2019} \right)}^2} + \left( {2018 - x} \right)\left( {x - 2019} \right)}} = \dfrac{{19}}{{49}}\\
 \to \dfrac{{1 - \left( {2018 - x} \right)\left( {x - 2019} \right)}}{{{{\left( {4037 - 2x} \right)}^2} + \left( {2018 - x} \right)\left( {x - 2019} \right)}} = \dfrac{{19}}{{49}}\\
 \to \dfrac{{1 - 2018x + 2018.2019 + {x^2} - 2019x}}{{4{x^2} - 2.2x.4037 + {{4037}^2} - {x^2} + 2019x + 2018x - 2018.2019}} = \dfrac{{19}}{{49}}\\
 \to \dfrac{{{x^2} - 4037x + 4074343}}{{3{x^2} - 12111x + 12223027}} = \dfrac{{19}}{{49}}\\
 \to 49{x^2} - 49.4037x + 49.4074343 = 57{x^2} - 19.12111x + 19.12223027\\
 \to 8{x^2} - 32296x + 32594706 = 0\\
 \to 4{x^2} - 16148x + 16297353 = 0\\
 \to {\left( {2x} \right)^2} - 2.2x.4037 + {\left( {4037} \right)^2} - 16 = 0\\
 \to {\left( {2x - 4037} \right)^2} = 16\\
 \to \left[ \begin{array}{l}
2x - 4037 = 4\\
2x - 4037 =  - 4
\end{array} \right.\\
 \to \left[ \begin{array}{l}
x = \dfrac{{4041}}{2}\\
x = \dfrac{{4033}}{2}
\end{array} \right.
\end{array}\)