Bài tập 4 trang 189 SGK Toán 12 NC

\(\begin{array}{*{20}{l}}
{\frac{1}{{2 - 3i}} = \frac{{2 + 3i}}{{4 - 9{i^2}}} = \frac{2}{{13}} + \frac{3}{{13}}i}\\
\begin{array}{l}
\frac{1}{{\frac{1}{2} - \frac{{\sqrt 3 }}{2}i}} = \frac{{\frac{1}{2} + \frac{{\sqrt 3 }}{2}i}}{{\frac{1}{4} - {{\left( {\frac{{\sqrt 3 }}{2}i} \right)}^2}}}\\
 = \frac{{\frac{1}{2} + \frac{{\sqrt 3 }}{2}i}}{1} = \frac{1}{2} + \frac{{\sqrt 3 }}{2}i
\end{array}\\
\begin{array}{l}
\frac{{3 - 2i}}{i} = \frac{{i(3 - 2i)}}{{{i^2}}} =  - i(3 - 2i)\\
 =  - 3i + 2{i^2} =  - 2 - 3i
\end{array}\\
\begin{array}{l}
\frac{{3 - 4i}}{{4 - i}} = \frac{{(3 - 4i)(4 + i)}}{{17}}\\
 = \frac{{16 - 13i}}{{17}} = \frac{{16}}{{17}} - \frac{{13}}{{17}}i.
\end{array}
\end{array}\)

-- Mod Toán 12 HỌC247