Câu hỏi 33

Ta có

\(\begin{array}{l}\left( {2 + i} \right)z + \dfrac{{2(1 + 2i)}}{{1 + i}} = 7 + 8i \Leftrightarrow \left( {2 + i} \right)z + \dfrac{{\left( {2 + 4i} \right)\left( {1 - i} \right)}}{{\left( {1 + i} \right)\left( {1 - i} \right)}} = 7 + 8i\\ \Leftrightarrow \left( {2 + i} \right)z + 3 + i = 7 + 8i \Leftrightarrow \left( {2 + i} \right)z = 4 + 7i\\ \Leftrightarrow z = \dfrac{{4 + 7i}}{{2 + i}} = \dfrac{{\left( {4 + 7i} \right)\left( {2 - i} \right)}}{{\left( {2 + i} \right)\left( {2 - i} \right)}} = 3 + 2i\end{array}\)

Suy ra \(w = z + 1 - 2i = 3 + 2i + 1 - 2i = 4 \Rightarrow \left| w \right| = 4.\)

Chọn D