Biết z1,z2 là 2 nghiệm của phương trình 3z^2 - 4z + 2 = 0

* Hướng dẫn giải

Đáp án A

$$\begin{gathered} \mathrm{Giải} \mathrm{PT}: 3{\mathrm{z}}^2-4\mathrm{z}+2=0\iff \left\{\begin{array}{l}{\mathrm{z}}_1=\frac{2}{3}+\frac{\sqrt{2}}{3}\mathrm{i}\\ {\mathrm{z}}_2=\frac{2}{3}-\frac{\sqrt{2}}{3}\mathrm{i}\end{array}\right. \\ \Rightarrow \mathrm{S}={\mathrm{z}}_1{\left(\overline{{\mathrm{z}}_2}\right)}^2+{\mathrm{z}}_2{\left(\overline{{\mathrm{z}}_1}\right)}^2 \\ =\left(\frac{2}{3}+\frac{\sqrt{2}}{3}\mathrm{i}\right){\left(\frac{2}{3}+\frac{\sqrt{2}}{3}\mathrm{i}\right)}^2+\left(\frac{2}{3}-\frac{\sqrt{2}}{3}\mathrm{i}\right){\left(\frac{2}{3}-\frac{\sqrt{2}}{3}\mathrm{i}\right)}^2 \\ ={\left(\frac{2}{3}+\frac{\sqrt{2}}{3}\mathrm{i}\right)}^3+{\left(\frac{2}{3}-\frac{\sqrt{2}}{3}\mathrm{i}\right)}^3 \\ =\frac{-8}{27} \end{gathered}$$