Cho số phức \(z = a + bi\,\,\left( {a,\,\,b \in \mathbb{R}} \right)\) thỏa mãn \(\left( {1 + i} \right)z + 2\bar z = 3 + 2i.\) Tính \(P = a + b.\)

Ta có \(\left( {1 + i} \right)z + 2\bar z = 3 + 2i \Leftrightarrow \left( {1 + i} \right)\left( {a + bi} \right) + 2\left( {a - bi} \right) = 3 + 2i\)

\[ \Leftrightarrow \left( {3a - b} \right) + \left( {a - b} \right)i = 3 + 2i\]\( \Leftrightarrow \left\{ \begin{array}{l}3a - b = 3\\a - b = 2\end{array} \right.\)

\( \Leftrightarrow \left\{ \begin{array}{l}3a - b = 3\\a - b = 2\end{array} \right.\)\( \Leftrightarrow \left\{ \begin{array}{l}3a - b = 3\\a - b = 2\end{array} \right.\)\( \Leftrightarrow \left\{ \begin{array}{l}a = \frac{1}{2}\\b =  - \frac{3}{2}\end{array} \right.\).

Suy ra \(P = a + b =  - 1\). Chọn D.