Câu hỏi 38

\(\begin{array}{l}{z^4} - 2{z^2} - 8 = 0 \Leftrightarrow \left( {{z^2} + 2} \right)({z^2} - 4) = 0\\ \Leftrightarrow \left[ \begin{array}{l}{z^2} + 2 = 0\\{z^2} - 4 = 0\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}{z^2} = - 2 = 2{i^2}\\{z^2} = 4\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}z = \pm i\sqrt 2 \\z = \pm 2\end{array} \right.\end{array}\)

Giả sử: \({z_1} = i\sqrt 2 ;{z_2} =  - i\sqrt 2 ;{z_3} = 2;{z_4} =  - 2\)

\( \Rightarrow P = \left| {{z_1}} \right|.\left| {{z_2}} \right|.\left| {{z_3}} \right|.\left| {{z_4}} \right| = \left| {i\sqrt 2 } \right|.\left| { - i\sqrt 2 } \right|.\left| 2 \right|.\left| { - 2} \right| = 8\)

Chọn B