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Câu hỏi: Xét hai số phức ${{z}_{1}},{{z}_{2}}$ thỏa mãn $\left| {{z}_{1}}+2{{z}_{2}} \right|=2$, $\left| 2{{z}_{1}}-3{{z}_{2}}-7i \right|=4$. Giá trị lớn nhất của biểu thức $P=\left| {{z}_{1}}-2i \right|+\left| {{z}_{2}}+i \right|$ bằng
A. $\dfrac{2\sqrt{3}}{3}$.
B. $2\sqrt{3}$.
C. $4\sqrt{3}$.
D. $\dfrac{4\sqrt{3}}{3}$.
Để ý ${{z}_{1}}+2{{z}_{2}}=\left( {{z}_{1}}-2i \right)+2\left( {{z}_{2}}+i \right)$ ; $2{{z}_{1}}-3{{z}_{2}}-7i=2\left( {{z}_{1}}-2i \right)-3\left( {{z}_{2}}+i \right)$.
Gọi $A\left( {{z}_{1}}-2i \right){{,}^{{}}}B\left( {{z}_{2}}+i \right)\Rightarrow \left\{ \begin{aligned}
& \left| {{z}_{1}}+2{{z}_{2}} \right|=2 \\
& \left| 2{{z}_{1}}-3{{z}_{2}}-7i \right|=4 \\
\end{aligned} \right. $ $ \Leftrightarrow \left\{ \begin{aligned}
& {{\left( \overrightarrow{OA}+2\overrightarrow{OB} \right)}^{2}}=4 \\
& {{\left( 2\overrightarrow{OA}-3\overrightarrow{OB} \right)}^{2}}=16 \\
\end{aligned} \right.$
$\Leftrightarrow \left\{ \begin{aligned}
& {{\overrightarrow{OA}}^{2}}+4{{\overrightarrow{OB}}^{2}}+4\overrightarrow{OA}.\overrightarrow{OB}={{4}^{{}}}\left( 1 \right) \\
& 4{{\overrightarrow{OA}}^{2}}+9{{\overrightarrow{OB}}^{2}}-12\overrightarrow{OA}\overrightarrow{OB}={{16}^{{}}}\left( 2 \right) \\
\end{aligned} \right.$.
Lấy $3\times \left( 1 \right)+\left( 2 \right)\Rightarrow 7O{{A}^{2}}+21O{{B}^{2}}=12+16=28\Leftrightarrow O{{A}^{2}}+3O{{B}^{2}}=4$.
Vì vậy $P=OA+OB=1.OA+\dfrac{1}{\sqrt{3}}.\sqrt{3}OB\le \sqrt{\left( 1+{{\left( \dfrac{1}{\sqrt{3}} \right)}^{2}} \right)\left( O{{A}^{2}}+3O{{B}^{2}} \right)}=\dfrac{4\sqrt{3}}{3}$.
A. $\dfrac{2\sqrt{3}}{3}$.
B. $2\sqrt{3}$.
C. $4\sqrt{3}$.
D. $\dfrac{4\sqrt{3}}{3}$.
Để ý ${{z}_{1}}+2{{z}_{2}}=\left( {{z}_{1}}-2i \right)+2\left( {{z}_{2}}+i \right)$ ; $2{{z}_{1}}-3{{z}_{2}}-7i=2\left( {{z}_{1}}-2i \right)-3\left( {{z}_{2}}+i \right)$.
Gọi $A\left( {{z}_{1}}-2i \right){{,}^{{}}}B\left( {{z}_{2}}+i \right)\Rightarrow \left\{ \begin{aligned}
& \left| {{z}_{1}}+2{{z}_{2}} \right|=2 \\
& \left| 2{{z}_{1}}-3{{z}_{2}}-7i \right|=4 \\
\end{aligned} \right. $ $ \Leftrightarrow \left\{ \begin{aligned}
& {{\left( \overrightarrow{OA}+2\overrightarrow{OB} \right)}^{2}}=4 \\
& {{\left( 2\overrightarrow{OA}-3\overrightarrow{OB} \right)}^{2}}=16 \\
\end{aligned} \right.$
$\Leftrightarrow \left\{ \begin{aligned}
& {{\overrightarrow{OA}}^{2}}+4{{\overrightarrow{OB}}^{2}}+4\overrightarrow{OA}.\overrightarrow{OB}={{4}^{{}}}\left( 1 \right) \\
& 4{{\overrightarrow{OA}}^{2}}+9{{\overrightarrow{OB}}^{2}}-12\overrightarrow{OA}\overrightarrow{OB}={{16}^{{}}}\left( 2 \right) \\
\end{aligned} \right.$.
Lấy $3\times \left( 1 \right)+\left( 2 \right)\Rightarrow 7O{{A}^{2}}+21O{{B}^{2}}=12+16=28\Leftrightarrow O{{A}^{2}}+3O{{B}^{2}}=4$.
Vì vậy $P=OA+OB=1.OA+\dfrac{1}{\sqrt{3}}.\sqrt{3}OB\le \sqrt{\left( 1+{{\left( \dfrac{1}{\sqrt{3}} \right)}^{2}} \right)\left( O{{A}^{2}}+3O{{B}^{2}} \right)}=\dfrac{4\sqrt{3}}{3}$.
Đáp án D.