Bài tập 4.19 trang 204 SBT Toán 12

a)

\(\begin{array}{l}
\frac{{(2 + i) + (1 + i)(4 - 3i)}}{{3 + 2i}} = \frac{{2 + i + 7 + i}}{{3 + 2i}}\\
 = \frac{{9 + 2i}}{{3 + 2i}}\\
 = \frac{{(9 + 2i)(3 - 2i)}}{{9 - 4}}\\
 = \frac{{31 - 12i}}{{13}} = \frac{{31}}{{13}} - \frac{{12}}{{13}}i
\end{array}\)

b) 

\(\begin{array}{l}
\frac{{(3 - 4i)(1 + 2i)}}{{1 - 2i}} + 4 - 3i\\
 = \frac{{11 + 2i}}{{1 - 2i}} + 4 - 3i\\
 = (\frac{{11 + 2i)(1 + 2i)}}{{1 + 4}} + 4 - 3i\\
 = \frac{{7 + 24i}}{5} + 4 - 3i\\
 = \frac{{27}}{5} + \frac{9}{5}i
\end{array}\)

-- Mod Toán 12 HỌC247