a) Ta có: \(\left( {2x + 1} \right) + \left( {1 - 2y} \right)i\) \( = \left( {2 - x} \right) + \left( {3y - 2} \right)i\)
\( \Leftrightarrow \left\{ \begin{array}{l}2x + 1 = 2 - x\\1 - 2y = 3y - 2\end{array} \right.\) \( \Leftrightarrow \left\{ \begin{array}{l}3x = 1\\5y = 3\end{array} \right.\) \( \Leftrightarrow \left\{ \begin{array}{l}x = \dfrac{1}{3}\\y = \dfrac{3}{5}\end{array} \right.\)
Vậy \(x = \dfrac{1}{3},y = \dfrac{3}{5}\)
b) \(4x + 3 + \left( {3y - 2} \right)i\) \( = {\rm{ }}y\; + 1 + \left( {x - 3} \right)i\)
\( \Leftrightarrow \left\{ \begin{array}{l}4x + 3 = y + 1\\3y - 2 = x - 3\end{array} \right.\) \( \Leftrightarrow \left\{ \begin{array}{l}4x - y = - 2\\x - 3y = 1\end{array} \right.\) \( \Leftrightarrow \left\{ {} \right.\)
Vậy \(x = - \dfrac{7}{{11}},y = - \dfrac{6}{{11}}\).
c) \(x + 2y + \left( {2x - y} \right)i\) \( = 2x + y + \left( {x + 2y} \right)\)
\( \Leftrightarrow \left\{ \begin{array}{l}x + 2y = 2x + y\\2x - y = x + 2y\end{array} \right.\) \( \Leftrightarrow \left\{ \begin{array}{l}x - y = 0\\x - 3y = 0\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}x = 0\\y = 0\end{array} \right.\)
Vậy \(x = y = 0\).
