Phương pháp giải:
Giải chi tiết:Kiến thức: \(\left\{ \begin{array}{l}\left| {{z_1} + {z_2}} \right| = 10\\left| {{z_1} - {z_2}} \right| = 2\end{array} \right. \Rightarrow {\left| {{z_1} + {z_2}} \right|^2} + {\left| {{z_1} - {z_2}} \right|^2} = 104 \Leftrightarrow 2\left( {{{\left| {{z_1}} \right|}^2} + {{\left| {{z_2}} \right|}^2}} \right) = 104\,\,\,\left( 1 \right)\)Theo BĐT Bunhia:\(P = 1.\left| {{z_1}} \right| + 1.\left| {{z_2}} \right| \le \sqrt {\left( {{1^2} + {1^2}} \right)\left( {{{\left| {{z_1}} \right|}^2} + {{\left| {{z_2}} \right|}^2}} \right)} \) \( \Leftrightarrow P \le \sqrt {2.\left( {{{\left| {{z_1}} \right|}^2} + {{\left| {{z_2}} \right|}^2}} \right)} \,\,\,\,\left( 2 \right)\)Từ \(\left( 1 \right),\left( 2 \right) \Rightarrow P \le \sqrt {104}  \Rightarrow {P_{\max }} = 2\sqrt {26} \)Max khi \(\left\{ \begin{array}{l}\left| {{z_1}} \right| = \left| {{z_2}} \right|\{z_1} + {z_2} = 8 + 6i\\left| {{z_1} - {z_2}} \right| = 2\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}\left| {{z_1}} \right| = \left| {8 + 6i - {z_1}} \right|\{z_1} - \left( {8 + 6i - {z_1}} \right) = 2\{z_2} = 8 + 6i - {z_1}\end{array} \right.\)