Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
1,\\
z = x + yi\\
\left( {2 - i} \right)\left( {3z + 1} \right) = \left( {z + 2} \right)\left( {4 - 5i} \right)\\
\Leftrightarrow 6z + 2 - 3iz - i = 4z - 5iz + 8 - 10i\\
\Leftrightarrow 2z + 2iz + 9i - 6 = 0\\
\Leftrightarrow 2\left( {x + yi} \right) + 2i\left( {x + yi} \right) + 9i - 6 = 0\\
\Leftrightarrow 2x + 2yi + 2xi + 2y{i^2} + 9i - 6 = 0\\
\Leftrightarrow \left( {2x - 2y - 6} \right) + \left( {2y + 2x + 9} \right)i = 0\,\,\,\,\,\,\,\,\left( {{i^2} = - 1} \right)\\
\Leftrightarrow \left\{ \begin{array}{l}
2x - 2y - 6 = 0\\
2x + 2y + 9 = 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x = - \frac{3}{4}\\
y = - \frac{{15}}{4}
\end{array} \right.\\
\Rightarrow \left| z \right| = \sqrt {{x^2} + {y^2}} = \frac{{3\sqrt {26} }}{4}\\
2,\\
{z^2} + 8z + 17 = 0\\
\Leftrightarrow \left( {{z^2} + 2.z.4 + {4^2}} \right) + 1 = 0\\
\Leftrightarrow {\left( {z + 4} \right)^2} = - 1\\
\Leftrightarrow {\left( {z + 4} \right)^2} = {i^2}\\
\Leftrightarrow \left[ \begin{array}{l}
z + 4 = i\\
z + 4 = - i
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
z = - 4 + i\\
z = - 4 - i
\end{array} \right.\\
\Rightarrow \frac{1}{{{z_1}}} + \frac{1}{{{z_2}}} = - \frac{8}{{17}}
\end{array}\)
