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Câu hỏi: Xét hai số phức ${{z}_{1}},{{z}_{2}}$ thỏa mãn $\left| {{z}_{1}}-2{{z}_{2}} \right|=3$ và $\left| 3{{z}_{1}}+{{z}_{2}} \right|=2$. Khi $\left| {{z}_{1}}-\sqrt{3}i.{{z}_{2}}+i \right|$ đạt giá trị nhỏ nhất thì $\left| {{z}_{1}}-{{z}_{2}} \right|$ bằng
A. $\dfrac{17\sqrt{2}}{7}$.
B. $\dfrac{2\sqrt{43}}{7}$.
C. $\dfrac{2\sqrt{31}}{7}$.
D. $\dfrac{\sqrt{170}}{7}$.
A. $\dfrac{17\sqrt{2}}{7}$.
B. $\dfrac{2\sqrt{43}}{7}$.
C. $\dfrac{2\sqrt{31}}{7}$.
D. $\dfrac{\sqrt{170}}{7}$.
Để cho đơn giản ta đặt
$a={{z}_{1}}-2{{x}_{2}}$ và $b=3{{z}_{1}}+{{z}_{2}}$ thì ${{z}_{1}}=\dfrac{a+2b}{7};{{z}_{2}}=\dfrac{b-3a}{7}$ và $\left| a \right|=3;\left| b \right|=2$
Khi đó $p=\left| {{z}_{1}}-\sqrt{3}i.{{z}_{2}}+i \right|=\left| \dfrac{a+2b}{7}-\sqrt{3}i\dfrac{b-3a}{7}+i \right|$ $=\dfrac{1}{7}\left| a\left( 1+3\sqrt{3}i \right)+b\left( 2-\sqrt{3}i \right)+7i \right|$
$\ge \dfrac{1}{7}\left[ \left| a\left( 1+3\sqrt{3}i \right)+b\left( 2-\sqrt{3}i \right) \right|-7i \right]$
$\ge \dfrac{1}{7}\left[ \left| a\left( 1+3\sqrt{3}i \right) \right|-\left| b\left( 2-\sqrt{3}i \right) \right|-\left| 7i \right| \right]$ $=\dfrac{1}{7}\left( 6\sqrt{7}-2\sqrt{7}-7 \right)=\dfrac{4\sqrt{7}-7}{7}$
Dấu bằng xãy ra khi $a\left( 1+3\sqrt{3}i \right)=-6\sqrt{7}i;b\left( 2-\sqrt{3}i \right)=2\sqrt{7}i$
$\Rightarrow \left| {{z}_{1}}-{{z}_{2}} \right|=\left| \dfrac{4a+b}{7} \right|=$ $\left| \dfrac{4\left( \dfrac{-6\sqrt{7}i}{1+3\sqrt{3}i} \right)+\dfrac{2\sqrt{7}i}{2-\sqrt{3}i}}{7} \right|=\dfrac{2\sqrt{43}}{7}$.
$a={{z}_{1}}-2{{x}_{2}}$ và $b=3{{z}_{1}}+{{z}_{2}}$ thì ${{z}_{1}}=\dfrac{a+2b}{7};{{z}_{2}}=\dfrac{b-3a}{7}$ và $\left| a \right|=3;\left| b \right|=2$
Khi đó $p=\left| {{z}_{1}}-\sqrt{3}i.{{z}_{2}}+i \right|=\left| \dfrac{a+2b}{7}-\sqrt{3}i\dfrac{b-3a}{7}+i \right|$ $=\dfrac{1}{7}\left| a\left( 1+3\sqrt{3}i \right)+b\left( 2-\sqrt{3}i \right)+7i \right|$
$\ge \dfrac{1}{7}\left[ \left| a\left( 1+3\sqrt{3}i \right)+b\left( 2-\sqrt{3}i \right) \right|-7i \right]$
$\ge \dfrac{1}{7}\left[ \left| a\left( 1+3\sqrt{3}i \right) \right|-\left| b\left( 2-\sqrt{3}i \right) \right|-\left| 7i \right| \right]$ $=\dfrac{1}{7}\left( 6\sqrt{7}-2\sqrt{7}-7 \right)=\dfrac{4\sqrt{7}-7}{7}$
Dấu bằng xãy ra khi $a\left( 1+3\sqrt{3}i \right)=-6\sqrt{7}i;b\left( 2-\sqrt{3}i \right)=2\sqrt{7}i$
$\Rightarrow \left| {{z}_{1}}-{{z}_{2}} \right|=\left| \dfrac{4a+b}{7} \right|=$ $\left| \dfrac{4\left( \dfrac{-6\sqrt{7}i}{1+3\sqrt{3}i} \right)+\dfrac{2\sqrt{7}i}{2-\sqrt{3}i}}{7} \right|=\dfrac{2\sqrt{43}}{7}$.
Đáp án B.