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Câu hỏi: Cho các số phức ${{z}_{1}}\not{=}0, {{z}_{2}}\not{=}0$ thỏa mãn điều kiện $\dfrac{2}{{{z}_{1}}}+\dfrac{1}{{{z}_{2}}}=\dfrac{1}{{{z}_{1}}+{{z}_{2}}}.$ Tính giá trị của biểu thức $P=\left| \dfrac{{{z}_{1}}}{{{z}_{2}}} \right|+\left| \dfrac{{{z}_{2}}}{{{z}_{1}}} \right|.$.
A. $\dfrac{1}{\sqrt{2}}$.
B. $P=2$.
C. $\dfrac{3\sqrt{2}}{2}$.
D. $\sqrt{2}$.
A. $\dfrac{1}{\sqrt{2}}$.
B. $P=2$.
C. $\dfrac{3\sqrt{2}}{2}$.
D. $\sqrt{2}$.
$\dfrac{2}{{{z}_{1}}}+\dfrac{1}{{{z}_{2}}}=\dfrac{1}{{{z}_{1}}+{{z}_{2}}}$ $\Leftrightarrow \dfrac{2{{z}_{2}}+{{z}_{1}}}{{{z}_{1}}{{z}_{2}}}=\dfrac{1}{{{z}_{1}}+{{z}_{2}}}$ $\Leftrightarrow \left( 2{{z}_{2}}+{{z}_{1}} \right)\left( {{z}_{1}}+{{z}_{2}} \right)-{{z}_{1}}{{z}_{2}}=0$ $\Leftrightarrow 2{{z}_{1}}{{z}_{2}}+2z_{2}^{2}+z_{1}^{2}+{{z}_{1}}{{z}_{2}}-{{z}_{1}}{{z}_{2}}=0$ $\Leftrightarrow 2{{z}_{1}}{{z}_{2}}+2z_{2}^{2}+z_{1}^{2}=0$ $\Leftrightarrow {{\left( \dfrac{{{z}_{1}}}{{{z}_{2}}} \right)}^{2}}+2\dfrac{{{z}_{1}}}{{{z}_{2}}}+2=0$ $\Leftrightarrow \left[ \begin{aligned}
& \dfrac{{{z}_{1}}}{{{z}_{2}}}=-1-i \\
& \dfrac{{{z}_{1}}}{{{z}_{2}}}=-1+i \\
\end{aligned} \right.$$\Rightarrow \left| \dfrac{{{z}_{1}}}{{{z}_{2}}} \right|=\sqrt{2} $; $ \left| \dfrac{{{z}_{2}}}{{{z}_{1}}} \right|=\dfrac{1}{\left| \dfrac{{{z}_{1}}}{{{z}_{2}}} \right|}=\dfrac{1}{\sqrt{2}} $ $ \Rightarrow P=\sqrt{2}+\dfrac{1}{\sqrt{2}}=\dfrac{3\sqrt{2}}{2}$.
& \dfrac{{{z}_{1}}}{{{z}_{2}}}=-1-i \\
& \dfrac{{{z}_{1}}}{{{z}_{2}}}=-1+i \\
\end{aligned} \right.$$\Rightarrow \left| \dfrac{{{z}_{1}}}{{{z}_{2}}} \right|=\sqrt{2} $; $ \left| \dfrac{{{z}_{2}}}{{{z}_{1}}} \right|=\dfrac{1}{\left| \dfrac{{{z}_{1}}}{{{z}_{2}}} \right|}=\dfrac{1}{\sqrt{2}} $ $ \Rightarrow P=\sqrt{2}+\dfrac{1}{\sqrt{2}}=\dfrac{3\sqrt{2}}{2}$.
Đáp án C.