Cho số phức \(z = a + bi\,\,\left( {a,\,b \in \mathbb{R}} \right)\) thỏa mãn \(\left( {2 + i} \right)\left( {\bar z + 1 - i} \right) - \left( {2 - 3i} \right)\left( {z + i} \right) = 2 + 5i\). Tính \(S = 2a - 3b\).

Ta có \(z = a + bi\,\,\left( {a,\,b \in \mathbb{R}} \right) \Rightarrow \bar z = a - bi\)

Khi đó \(\left( {2 + i} \right)\left( {\bar z + 1 - i} \right) - \left( {2 - 3i} \right)\left( {z + i} \right) = 2 + 5i\)

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

\[ \Leftrightarrow \left( {2 + i} \right)\left[ {a + 1 - \left( {b + 1} \right)i} \right] - \left( {2 - 3i} \right)\left[ {a + \left( {b + 1} \right)i} \right] = 2 + 5i\]

\( \Leftrightarrow 2\left( {a + 1} \right) - 2\left( {b + 1} \right)i + \left( {a + 1} \right)i - \left( {b + 1} \right){i^2} - \left[ {2a + 2\left( {b + 1} \right)i - 3ai - 3\left( {b + 1} \right){i^2}} \right] = 2 + 5i\)

\( \Leftrightarrow 2a + 2 - \left( {2b + 2} \right)i + \left( {a + 1} \right)i + b + 1 - 2a - \left( {2b + 2} \right)i + 3ai - 3b - 3 = 2 + 5i\)

\( \Leftrightarrow  - 2b + \left( {4a - 4b - 3} \right)i = 2 + 5i \Leftrightarrow \left\{ {\begin{array}{*{20}{l}}{ - 2b = 2}\\{4a - 4b - 3 = 5}\end{array} \Leftrightarrow \left\{ {\begin{array}{*{20}{l}}{a = 1}\\{b =  - 1}\end{array}} \right.} \right.\).

Do đó \({\rm{S}} = 2{\rm{a}} - 3\;{\rm{b}} = 2 \cdot 1 - 3 \cdot \left( { - 1} \right) = 5\).

Đáp án: 5.