Đáp án:
\[\left| {{z_1} + {z_2}} \right| = \sqrt 2 \]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
{z_1} = {x_1} + {y_1}i\\
{z_2} = {x_2} + {y_2}i\\
\left\{ \begin{array}{l}
\left| {{z_1}} \right| = \left| {{z_2}} \right| = \sqrt {13} \\
\left| {{z_1} - {z_2}} \right| = 5\sqrt 2
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
\sqrt {{x_1}^2 + {y_1}^2} = \sqrt {{x_2}^2 + {y_2}^2} = \sqrt {13} \\
\left| {\left( {{x_1} + {y_1}i} \right) - \left( {{x_2} + {y_2}i} \right)} \right| = 5\sqrt 2
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
{x_1}^2 + {y_1}^2 = {x_2}^2 + {y_2}^2 = 13\\
{\left( {{x_1} - {x_2}} \right)^2} + {\left( {{y_1} - {y_2}} \right)^2} = 50
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
{x_1}^2 + {y_1}^2 = {x_2}^2 + {y_2}^2 = 13\\
{x_1}^2 - 2{x_1}{x_2} + {x_2}^2 + {y_1}^2 - 2{y_1}{y_2} + {y_2}^2 = 50
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
{x_1}^2 + {y_1}^2 = {x_2}^2 + {y_2}^2 = 13\\
{x_1}{x_2} + {y_1}{y_2} = - 12
\end{array} \right.\\
A = \left| {{z_1} + {z_2}} \right| = \left| {{x_1} + {y_1}i + {x_2} + {y_2}i} \right|\\
= \left| {\left( {{x_1} + {x_2}} \right) + \left( {{y_1} + {y_2}} \right)i} \right|\\
= \sqrt {{{\left( {{x_1} + {x_2}} \right)}^2} + {{\left( {{y_1} + {y_2}} \right)}^2}} \\
= \sqrt {\left( {{x_1}^2 + {y_1}^2} \right) + \left( {{x_2}^2 + {y_2}^2} \right) + 2\left( {{x_1}{x_2} + {y_1}{y_2}} \right)} \\
= \sqrt {13 + 13 + 2.\left( { - 12} \right)} = \sqrt 2
\end{array}\)
Vậy \(\left| {{z_1} + {z_2}} \right| = \sqrt 2 \)
