Gọi z1, z2,z3, z4 là các nghiệm phức của phương trình ${(z_{}^2+z)}_{}^2+4(z_{}^2+z)-12=0$. Tính $S={\left|z_1^{}\right|}_{}^2+{\left|z_2^{}\right|}_{}^2+{\left|z_3^{}\right|}_{}^2+{\left|z_4^{}\right|}_{}^2$

A. S = 18

Ta có ${(z_{}^2+z)}_{}^2+4(z_{}^2+z)-12=0\iff \left[\begin{array}{l}{z}_{}^2+z=2\\ z_{}^2+z=-6\end{array}\right.\iff \left[\begin{array}{l}{z}_{}^2+z-2=0\\ z_{}^2+z+6=0\end{array}\right.\iff \left[\begin{array}{l}z=1\\ z=-2\\ z=\frac{-1+\sqrt{23}i}{2}\\ z=\frac{-1-\sqrt{23}i}{2}\end{array}\right.$

$S={\left|z_1^{}\right|}_{}^2+{\left|z_2^{}\right|}_{}^2+{\left|z_3^{}\right|}_{}^2+{\left|z_4^{}\right|}_{}^2=1_{}^2+2_{}^2+\frac{1+23}{4}.2=17$

Chọn: C