Đáp án:
\(\left[ \begin{array}{l}
x = 1\\
x = 2\\
x = 0
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
2018.\left| {x - 1} \right| + {\left( {x - 1} \right)^2} = 2019.\left| {1 - x} \right|\\
\to {\left( {x - 1} \right)^2} = 2019.\left| {x - 1} \right| - 2018.\left| {x - 1} \right|\\
\to {\left( {x - 1} \right)^2} = \left| {x - 1} \right|\\
\to \left[ \begin{array}{l}
{\left( {x - 1} \right)^2} = \left( {x - 1} \right)\\
{\left( {x - 1} \right)^2} = - \left( {x - 1} \right)
\end{array} \right.\\
\to \left[ \begin{array}{l}
\left( {x - 1} \right)\left( {x - 1 - 1} \right) = 0\\
\left( {x - 1} \right)\left( {x - 1 + 1} \right) = 0
\end{array} \right.\\
\to \left[ \begin{array}{l}
x - 1 = 0\\
x - 2 = 0\\
x = 0
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 1\\
x = 2\\
x = 0
\end{array} \right.
\end{array}\)
