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Câu hỏi: Gọi ${{z}_{1}},{{z}_{2}}$ là các nghiệm phức của phương trình ${{z}^{2}}+z+1=0$, đặt $\text{w}=z_{1}^{2021}+z_{2}^{2021}.$ Khi đó
A. $$ /B]text{w}={{2}^{2021}}. $
B. $ \text{w}=-1. $
C. $ \text{w}={{2}^{2021}}i. $
D. $ \text{w}=1.$
A. $$ /B]text{w}={{2}^{2021}}. $
B. $ \text{w}=-1. $
C. $ \text{w}={{2}^{2021}}i. $
D. $ \text{w}=1.$
Ta có:
${{z}^{2}}+z+1=0\Leftrightarrow \left[ \begin{aligned}
& {{z}_{1}}=\dfrac{-1+\sqrt{3}i}{2} \\
& {{z}_{2}}=\dfrac{-1-\sqrt{3}i}{2} \\
\end{aligned} \right.$
${{z}_{1}}=\dfrac{-1+\sqrt{3}i}{2}\Rightarrow z_{1}^{3}=1\Rightarrow {{\left( z_{1}^{3} \right)}^{673}}={{1}^{673}}\Rightarrow z_{1}^{2019}=1\Rightarrow z_{1}^{2021}=z_{1}^{2}=\dfrac{-1-\sqrt{3}i}{2}$
${{z}_{2}}=\dfrac{-1-\sqrt{3}i}{2}\Rightarrow z_{2}^{3}=1\Rightarrow {{\left( z_{2}^{3} \right)}^{673}}={{1}^{673}}\Rightarrow z_{2}^{2019}=1\Rightarrow z_{2}^{2021}=z_{2}^{2}=\dfrac{-1+\sqrt{3}i}{2}$
$\text{w}=z_{1}^{2021}+z_{2}^{2021}=\dfrac{-1-\sqrt{3}i}{2}+\dfrac{-1+\sqrt{3}i}{2}=-1$.
${{z}^{2}}+z+1=0\Leftrightarrow \left[ \begin{aligned}
& {{z}_{1}}=\dfrac{-1+\sqrt{3}i}{2} \\
& {{z}_{2}}=\dfrac{-1-\sqrt{3}i}{2} \\
\end{aligned} \right.$
${{z}_{1}}=\dfrac{-1+\sqrt{3}i}{2}\Rightarrow z_{1}^{3}=1\Rightarrow {{\left( z_{1}^{3} \right)}^{673}}={{1}^{673}}\Rightarrow z_{1}^{2019}=1\Rightarrow z_{1}^{2021}=z_{1}^{2}=\dfrac{-1-\sqrt{3}i}{2}$
${{z}_{2}}=\dfrac{-1-\sqrt{3}i}{2}\Rightarrow z_{2}^{3}=1\Rightarrow {{\left( z_{2}^{3} \right)}^{673}}={{1}^{673}}\Rightarrow z_{2}^{2019}=1\Rightarrow z_{2}^{2021}=z_{2}^{2}=\dfrac{-1+\sqrt{3}i}{2}$
$\text{w}=z_{1}^{2021}+z_{2}^{2021}=\dfrac{-1-\sqrt{3}i}{2}+\dfrac{-1+\sqrt{3}i}{2}=-1$.
Đáp án B.