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Câu hỏi: Giả sử ${{z}_{1}}, {{z}_{2}}$ là hai nghiệm phức của phương trình $\left| \left( 2+i \right)\left| z \right|z-\left( 1-2i \right)z \right|=\left| 1+3i \right|$ và $\left| {{z}_{1}}-{{z}_{2}} \right|=1$. Tính $M=\left| 2{{z}_{1}}+3{{z}_{2}} \right|$.
A. $M=5$.
B. $M=\sqrt{19}$.
C. $M=19$.
D. $M=25$.
A. $M=5$.
B. $M=\sqrt{19}$.
C. $M=19$.
D. $M=25$.
$\left| \left( 2+i \right)\left| z \right|z-\left( 1-2i \right)z \right|=\left| 1+3i \right|\Leftrightarrow \left| z\left[ \left( 2\left| z \right|-1 \right)+\left( \left| z \right|+2 \right)i \right] \right|=\sqrt{10}$
$\Leftrightarrow \left| z \right|\sqrt{{{\left( 2\left| z \right|-1 \right)}^{2}}+{{\left( \left| z \right|+2 \right)}^{2}}}=\sqrt{10}\Leftrightarrow 5{{\left| z \right|}^{4}}+5{{\left| z \right|}^{2}}-10=0\Leftrightarrow {{\left| z \right|}^{2}}=1\Leftrightarrow \left| z \right|=1$
Gọi ${{z}_{1}}={{a}_{1}}+{{b}_{1}}i, {{z}_{2}}={{a}_{2}}+{{b}_{2}}i$.
Ta có: $\left| {{z}_{1}} \right|=\left| {{z}_{2}} \right|=1\Rightarrow {{a}_{1}}^{2}+{{b}_{1}}^{2}={{a}_{2}}^{2}+{{b}_{2}}^{2}=1$
Ta có: $\left| {{z}_{1}}-{{z}_{2}} \right|=1\Rightarrow {{\left( {{a}_{1}}-{{a}_{2}} \right)}^{2}}+{{\left( {{b}_{1}}-{{b}_{2}} \right)}^{2}}=1\Rightarrow {{a}_{1}}{{a}_{2}}+{{b}_{1}}{{b}_{2}}=\dfrac{1}{2}$
Ta có: $M=\left| 2{{z}_{1}}+3{{z}_{2}} \right|=\left| \left( 2{{a}_{1}}+3{{a}_{2}} \right)+\left( 2{{b}_{1}}+3{{b}_{2}} \right)i \right|=\sqrt{{{\left( 2{{a}_{1}}+3{{a}_{2}} \right)}^{2}}+{{\left( 2{{b}_{1}}+3{{b}_{2}} \right)}^{2}}}$
$=\sqrt{4\left( {{a}_{1}}^{2}+{{b}_{1}}^{2} \right)+12\left( {{a}_{1}}{{a}_{2}}+{{b}_{1}}{{b}_{2}} \right)+9\left( {{a}_{2}}^{2}+{{b}_{2}}^{2} \right)}=\sqrt{19}$.
$\Leftrightarrow \left| z \right|\sqrt{{{\left( 2\left| z \right|-1 \right)}^{2}}+{{\left( \left| z \right|+2 \right)}^{2}}}=\sqrt{10}\Leftrightarrow 5{{\left| z \right|}^{4}}+5{{\left| z \right|}^{2}}-10=0\Leftrightarrow {{\left| z \right|}^{2}}=1\Leftrightarrow \left| z \right|=1$
Gọi ${{z}_{1}}={{a}_{1}}+{{b}_{1}}i, {{z}_{2}}={{a}_{2}}+{{b}_{2}}i$.
Ta có: $\left| {{z}_{1}} \right|=\left| {{z}_{2}} \right|=1\Rightarrow {{a}_{1}}^{2}+{{b}_{1}}^{2}={{a}_{2}}^{2}+{{b}_{2}}^{2}=1$
Ta có: $\left| {{z}_{1}}-{{z}_{2}} \right|=1\Rightarrow {{\left( {{a}_{1}}-{{a}_{2}} \right)}^{2}}+{{\left( {{b}_{1}}-{{b}_{2}} \right)}^{2}}=1\Rightarrow {{a}_{1}}{{a}_{2}}+{{b}_{1}}{{b}_{2}}=\dfrac{1}{2}$
Ta có: $M=\left| 2{{z}_{1}}+3{{z}_{2}} \right|=\left| \left( 2{{a}_{1}}+3{{a}_{2}} \right)+\left( 2{{b}_{1}}+3{{b}_{2}} \right)i \right|=\sqrt{{{\left( 2{{a}_{1}}+3{{a}_{2}} \right)}^{2}}+{{\left( 2{{b}_{1}}+3{{b}_{2}} \right)}^{2}}}$
$=\sqrt{4\left( {{a}_{1}}^{2}+{{b}_{1}}^{2} \right)+12\left( {{a}_{1}}{{a}_{2}}+{{b}_{1}}{{b}_{2}} \right)+9\left( {{a}_{2}}^{2}+{{b}_{2}}^{2} \right)}=\sqrt{19}$.
Đáp án B.