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Câu hỏi: Cho số phức $m$ thỏa mãn $\left( 2-i \right)z-\left( 2+i \right)\overline{z}=2i$. Giá trị nhỏ nhất của ${{z}_{1}},{{z}_{2}}$ bằng:
A. $\dfrac{185}{63}$.
B. $\dfrac{11}{9}$.
C. $0$.
D. $2\left( {{z}_{1}}\left| {{z}_{2}} \right|+{{z}_{2}}\left| {{z}_{1}} \right| \right)=\left| {{z}_{1}}{{z}_{2}} \right|.$.
A. $\dfrac{185}{63}$.
B. $\dfrac{11}{9}$.
C. $0$.
D. $2\left( {{z}_{1}}\left| {{z}_{2}} \right|+{{z}_{2}}\left| {{z}_{1}} \right| \right)=\left| {{z}_{1}}{{z}_{2}} \right|.$.
Gọi $z=x+yi;$ $\overline{z}=x-yi$
$\left( 2-i \right)z-\left( 2+i \right)\overline{z}=2i$ $\Leftrightarrow \left( 2-i \right)\left( x+yi \right)-\left( 2+i \right)\left( x-yi \right)=2i$
$\Leftrightarrow 2x+2yi-xi+y-\left( 2x-2yi+xi+y \right)=2i$
$\Leftrightarrow \left\{ \begin{aligned}
& 2x+y-2x-y=0 \\
& 2y-x+2y-x=2 \\
\end{aligned} \right.\Leftrightarrow 2x-4y=-2\Leftrightarrow y=\dfrac{1}{2}x+\dfrac{1}{2}$
$\left| z \right|=\sqrt{{{x}^{2}}+{{y}^{2}}}=\sqrt{{{x}^{2}}+\dfrac{1}{4}{{x}^{2}}+\dfrac{1}{2}x+\dfrac{1}{4}}=\sqrt{\dfrac{5}{4}{{x}^{2}}+\dfrac{1}{2}x+\dfrac{1}{4}}$
Xét: $f\left( x \right)=\dfrac{5}{4}{{x}^{2}}+\dfrac{1}{2}x+\dfrac{1}{4}\Rightarrow f'\left( x \right)=\dfrac{5}{2}x+\dfrac{1}{2}=0\Rightarrow x=-\dfrac{1}{5}$
Dễ thấy hàm số nhỏ nhất tại $x=-\dfrac{1}{5}$ $\Rightarrow {{\left| z \right|}_{\min }}=\sqrt{\dfrac{5}{4}{{x}^{2}}+\dfrac{1}{2}x+\dfrac{1}{4}}=\dfrac{\sqrt{5}}{5}$.
$\left( 2-i \right)z-\left( 2+i \right)\overline{z}=2i$ $\Leftrightarrow \left( 2-i \right)\left( x+yi \right)-\left( 2+i \right)\left( x-yi \right)=2i$
$\Leftrightarrow 2x+2yi-xi+y-\left( 2x-2yi+xi+y \right)=2i$
$\Leftrightarrow \left\{ \begin{aligned}
& 2x+y-2x-y=0 \\
& 2y-x+2y-x=2 \\
\end{aligned} \right.\Leftrightarrow 2x-4y=-2\Leftrightarrow y=\dfrac{1}{2}x+\dfrac{1}{2}$
$\left| z \right|=\sqrt{{{x}^{2}}+{{y}^{2}}}=\sqrt{{{x}^{2}}+\dfrac{1}{4}{{x}^{2}}+\dfrac{1}{2}x+\dfrac{1}{4}}=\sqrt{\dfrac{5}{4}{{x}^{2}}+\dfrac{1}{2}x+\dfrac{1}{4}}$
Xét: $f\left( x \right)=\dfrac{5}{4}{{x}^{2}}+\dfrac{1}{2}x+\dfrac{1}{4}\Rightarrow f'\left( x \right)=\dfrac{5}{2}x+\dfrac{1}{2}=0\Rightarrow x=-\dfrac{1}{5}$
Dễ thấy hàm số nhỏ nhất tại $x=-\dfrac{1}{5}$ $\Rightarrow {{\left| z \right|}_{\min }}=\sqrt{\dfrac{5}{4}{{x}^{2}}+\dfrac{1}{2}x+\dfrac{1}{4}}=\dfrac{\sqrt{5}}{5}$.
Đáp án B.