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Câu hỏi: Cho ${{z}_{1}}=1+\sqrt{3}i;{{z}_{2}}=\dfrac{7+i}{4-3i};{{z}_{3}}=1-i.$
Tính giá trị biểu thức của $w=z_{1}^{25}.z_{2}^{10}.z_{3}^{2016}.$
A. ${{2}^{1037}}-{{2}^{1037}}\sqrt{3}i.$
B. $-{{2}^{1037}}\sqrt{3}+{{2}^{1037}}i.$
C. $-{{2}^{1021}}\sqrt{3}+{{2}^{1021}}i.$
D. ${{2}^{1021}}\sqrt{3}-{{2}^{1021}}i.$
Tính giá trị biểu thức của $w=z_{1}^{25}.z_{2}^{10}.z_{3}^{2016}.$
A. ${{2}^{1037}}-{{2}^{1037}}\sqrt{3}i.$
B. $-{{2}^{1037}}\sqrt{3}+{{2}^{1037}}i.$
C. $-{{2}^{1021}}\sqrt{3}+{{2}^{1021}}i.$
D. ${{2}^{1021}}\sqrt{3}-{{2}^{1021}}i.$
Ta có
$\left. \begin{aligned}
& z_{1}^{25}={{\left( 1+\sqrt{3}i \right)}^{25}}={{8}^{8}}+{{8}^{8}}\sqrt{3}i \\
& z_{2}^{10}={{\left( \dfrac{7+i}{4-3i} \right)}^{10}}={{\left( 2i \right)}^{5}}={{2}^{5}}i \\
& z_{3}^{2016}={{\left( 1-i \right)}^{2016}}={{\left( -2i \right)}^{1008}}={{2}^{1008}} \\
\end{aligned} \right\}\Rightarrow w=z_{1}^{25}.z_{2}^{10}.z_{3}^{2016}=-{{2}^{1037}}\sqrt{3}+{{2}^{1037}}i.$
$\left. \begin{aligned}
& z_{1}^{25}={{\left( 1+\sqrt{3}i \right)}^{25}}={{8}^{8}}+{{8}^{8}}\sqrt{3}i \\
& z_{2}^{10}={{\left( \dfrac{7+i}{4-3i} \right)}^{10}}={{\left( 2i \right)}^{5}}={{2}^{5}}i \\
& z_{3}^{2016}={{\left( 1-i \right)}^{2016}}={{\left( -2i \right)}^{1008}}={{2}^{1008}} \\
\end{aligned} \right\}\Rightarrow w=z_{1}^{25}.z_{2}^{10}.z_{3}^{2016}=-{{2}^{1037}}\sqrt{3}+{{2}^{1037}}i.$
Đáp án B.