Ta có \({x^2} - x = 4{y^2} - 2y \Leftrightarrow {x^2} - 4{y^2} = x - 2y\)

\( \Leftrightarrow \left( {x - 2y} \right)\left( {x + 2y} \right) = x - 2y \Leftrightarrow \left[ {\begin{array}{*{20}{l}}{x - 2y = 0}\\{x + 2y = 1}\end{array}} \right.\).

Khi đó, ta xét hai trường hợp sau:

TH1: \(\left\{ \begin{array}{l}x + \left| y \right| = 5\\x - 2y = 0\end{array} \right.\)\( \Leftrightarrow \left\{ \begin{array}{l}x = 2y\\x + \left| y \right| = 5\end{array} \right.\)\( \Leftrightarrow \left\{ \begin{array}{l}x = 2y\\2y + \left| y \right| = 5\end{array} \right.\)

\( \Leftrightarrow \left\{ \begin{array}{l}x = 2y\\\left| y \right| = 5 - 2y \ge 0\end{array} \right.\)\( \Leftrightarrow \left\{ \begin{array}{l}x = 2y\\y = 5 - 2y\end{array} \right.\)\( \Leftrightarrow \left( {x\,;\,\,y} \right) = \left( {\frac{{10}}{3}\,;\,\,\frac{5}{3}} \right)\)

TH2: \(\left\{ \begin{array}{l}x + \left| y \right| = 5\\x + 2y = 1\end{array} \right.\)\( \Leftrightarrow \left\{ \begin{array}{l}x = 1 - 2y\\x + \left| y \right| = 5\end{array} \right.\)\( \Leftrightarrow \left\{ \begin{array}{l}x = 1 - 2y\\1 - 2y + \left| y \right| = 5\end{array} \right.\)

\( \Leftrightarrow \left\{ \begin{array}{l}x = 1 - 2y\\\left| y \right| = 4 + 2y \ge 0\end{array} \right.\)\( \Leftrightarrow \left\{ \begin{array}{l}x = 1 - 2y\\y = 4 + 2y\end{array} \right.\)\( \Leftrightarrow \left( {x\,;\,\,y} \right) = \left( {\frac{{11}}{3}\,;\,\, - \frac{4}{3}} \right).\)

Chọn D.