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Câu hỏi: Gọi $z$ là một nghiệm của phương trình $z^{2}-z+1=0$. Giá trị của biểu thức $M=z^{2019}+z^{2018}+\dfrac{1}{z^{2019}}+\dfrac{1}{z^{2018}}+5$ bằng
A. $5$.
B. $2$.
C. $7$.
D. $-1$.
A. $5$.
B. $2$.
C. $7$.
D. $-1$.
${{z}^{2}}-z+1=0\Leftrightarrow \left[ \begin{aligned}
& z=\dfrac{1}{2}+\dfrac{\sqrt{3}}{2}i \\
& z=\dfrac{1}{2}-\dfrac{\sqrt{3}}{2}i \\
\end{aligned} \right.$.
Chọn $z=\dfrac{1}{2}+\dfrac{\sqrt{3}}{2}i$
Ta có:
${{z}^{2019}}={{\left( \dfrac{1}{2}+\dfrac{\sqrt{3}}{2}i \right)}^{2019}}={{\left[ {{\left( \dfrac{1}{2}+\dfrac{\sqrt{3}}{2}i \right)}^{3}} \right]}^{673}}={{\left( -1 \right)}^{673}}=-1$
${{z}^{2018}}={{\left( \dfrac{1}{2}+\dfrac{\sqrt{3}}{2}i \right)}^{2019-1}}$ $={{\left[ {{\left( \dfrac{1}{2}+\dfrac{\sqrt{3}}{2}i \right)}^{3}} \right]}^{673}}:\left( \dfrac{1}{2}+\dfrac{\sqrt{3}}{2}i \right)$
$={{\left( -1 \right)}^{673}}:\left( \dfrac{1}{2}+\dfrac{\sqrt{3}}{2}i \right)=-1:\left( \dfrac{1}{2}+\dfrac{\sqrt{3}}{2}i \right)=-\dfrac{1}{2}+\dfrac{\sqrt{3}}{2}i$.
$M={{z}^{2019}}+{{z}^{2018}}+\dfrac{1}{{{z}^{2019}}}+\dfrac{1}{{{z}^{2018}}}+5=-1+\left( -\dfrac{1}{2}+\dfrac{\sqrt{3}}{2}i \right)-1+\dfrac{1}{-\dfrac{1}{2}+\dfrac{\sqrt{3}}{2}i}+5=2$.
& z=\dfrac{1}{2}+\dfrac{\sqrt{3}}{2}i \\
& z=\dfrac{1}{2}-\dfrac{\sqrt{3}}{2}i \\
\end{aligned} \right.$.
Chọn $z=\dfrac{1}{2}+\dfrac{\sqrt{3}}{2}i$
Ta có:
${{z}^{2019}}={{\left( \dfrac{1}{2}+\dfrac{\sqrt{3}}{2}i \right)}^{2019}}={{\left[ {{\left( \dfrac{1}{2}+\dfrac{\sqrt{3}}{2}i \right)}^{3}} \right]}^{673}}={{\left( -1 \right)}^{673}}=-1$
${{z}^{2018}}={{\left( \dfrac{1}{2}+\dfrac{\sqrt{3}}{2}i \right)}^{2019-1}}$ $={{\left[ {{\left( \dfrac{1}{2}+\dfrac{\sqrt{3}}{2}i \right)}^{3}} \right]}^{673}}:\left( \dfrac{1}{2}+\dfrac{\sqrt{3}}{2}i \right)$
$={{\left( -1 \right)}^{673}}:\left( \dfrac{1}{2}+\dfrac{\sqrt{3}}{2}i \right)=-1:\left( \dfrac{1}{2}+\dfrac{\sqrt{3}}{2}i \right)=-\dfrac{1}{2}+\dfrac{\sqrt{3}}{2}i$.
$M={{z}^{2019}}+{{z}^{2018}}+\dfrac{1}{{{z}^{2019}}}+\dfrac{1}{{{z}^{2018}}}+5=-1+\left( -\dfrac{1}{2}+\dfrac{\sqrt{3}}{2}i \right)-1+\dfrac{1}{-\dfrac{1}{2}+\dfrac{\sqrt{3}}{2}i}+5=2$.
Đáp án B.