Google
Câu hỏi: Cho ba số phức ${{z}_{1}},{{z}_{2}},{{z}_{3}}$ thoả mãn $\left| {{z}_{1}} \right|=1, \left| {{z}_{2}} \right|=\sqrt{7}, \left| {{z}_{1}}-{{z}_{2}} \right|=\sqrt{2}$ và giá trị lớn nhất của $\left| 3{{z}_{1}}+2{{z}_{2}}+{{z}_{3}} \right|$ bằng 78. Giá trị $\left| {{z}_{3}} \right|$ bằng
A. $78-\sqrt{53}$.
B. $25$.
C. $78-\sqrt{73}$.
D. $5$.
A. $78-\sqrt{53}$.
B. $25$.
C. $78-\sqrt{73}$.
D. $5$.
Gọi ${{w}_{1}}=a+bi; {{w}_{2}}=c+di;\ a,b,c,d\in \mathbb{R}$.
Ta có $\left( {{a}^{2}}+{{b}^{2}} \right)\left( {{c}^{2}}+{{d}^{2}} \right)\ge {{\left( ac+bd \right)}^{2}}\Rightarrow 2\sqrt{\left( {{a}^{2}}+{{b}^{2}} \right)\left( {{c}^{2}}+{{d}^{2}} \right)}\ge 2\left( ac+bd \right)$
$\Leftrightarrow \left( {{a}^{2}}+{{b}^{2}} \right)+2\sqrt{\left( {{a}^{2}}+{{b}^{2}} \right)\left( {{c}^{2}}+{{d}^{2}} \right)}+\left( {{c}^{2}}+{{d}^{2}} \right)\ge \left( {{a}^{2}}+{{b}^{2}}+2ab \right)+\left( {{c}^{2}}+{{d}^{2}}+2cd \right)$
$\Leftrightarrow {{\left( \sqrt{{{a}^{2}}+{{b}^{2}}}+\sqrt{{{c}^{2}}+{{d}^{2}}} \right)}^{2}}\ge {{\left( a+b \right)}^{2}}+{{\left( c+d \right)}^{2}}$ $\Rightarrow \sqrt{{{a}^{2}}+{{b}^{2}}}+\sqrt{{{c}^{2}}+{{d}^{2}}}\ge \sqrt{{{\left( a+b \right)}^{2}}+{{\left( c+d \right)}^{2}}}$
Hay $\left| {{w}_{1}}+{{w}_{2}} \right|\le \left| {{w}_{1}} \right|+\left| {{w}_{2}} \right|$ (*)
Giả sử ${{z}_{1}}={{a}_{1}}+{{b}_{1}}i; {{z}_{2}}={{c}_{1}}+{{d}_{1}}i;\ {{a}_{1}},{{b}_{1}},{{c}_{1}},{{d}_{1}}\in \mathbb{R}$, khi đó $a_{1}^{2}+b_{1}^{2}=1; c_{1}^{2}+d_{1}^{2}=7$.
Ta có $2={{\left| {{z}_{1}}-{{z}_{2}} \right|}^{2}}={{\left( {{a}_{1}}-{{c}_{1}} \right)}^{2}}+{{\left( {{b}_{1}}-{{d}_{1}} \right)}^{2}}=8-2\left( {{a}_{1}}{{c}_{1}}+{{b}_{1}}{{d}_{1}} \right)\Rightarrow {{a}_{1}}{{c}_{1}}+{{b}_{1}}{{d}_{1}}=3$
Mặt khác ${{\left| 3{{z}_{1}}+2{{z}_{2}} \right|}^{2}}={{\left| 3\left( {{a}_{1}}+{{b}_{1}}i \right)+2\left( {{c}_{1}}+{{d}_{1}}i \right) \right|}^{2}}={{\left( 3{{a}_{1}}+2{{c}_{1}} \right)}^{2}}+{{\left( 3{{b}_{1}}+2{{d}_{1}} \right)}^{2}}$
$=9\left( a_{1}^{2}+b_{1}^{2} \right)+4\left( c_{1}^{2}+d_{1}^{2} \right)+12\left( {{a}_{1}}{{c}_{1}}+{{b}_{1}}{{d}_{1}} \right)=73$.
Theo bất đẳng thức (*) ta có $\left| 3{{z}_{1}}+2{{z}_{2}}+{{z}_{3}} \right|\le \left| 3{{z}_{1}}+2{{z}_{2}} \right|+\left| {{z}_{3}} \right|=78\Rightarrow \left| {{z}_{3}} \right|=78-\sqrt{73}$.
Ta có $\left( {{a}^{2}}+{{b}^{2}} \right)\left( {{c}^{2}}+{{d}^{2}} \right)\ge {{\left( ac+bd \right)}^{2}}\Rightarrow 2\sqrt{\left( {{a}^{2}}+{{b}^{2}} \right)\left( {{c}^{2}}+{{d}^{2}} \right)}\ge 2\left( ac+bd \right)$
$\Leftrightarrow \left( {{a}^{2}}+{{b}^{2}} \right)+2\sqrt{\left( {{a}^{2}}+{{b}^{2}} \right)\left( {{c}^{2}}+{{d}^{2}} \right)}+\left( {{c}^{2}}+{{d}^{2}} \right)\ge \left( {{a}^{2}}+{{b}^{2}}+2ab \right)+\left( {{c}^{2}}+{{d}^{2}}+2cd \right)$
$\Leftrightarrow {{\left( \sqrt{{{a}^{2}}+{{b}^{2}}}+\sqrt{{{c}^{2}}+{{d}^{2}}} \right)}^{2}}\ge {{\left( a+b \right)}^{2}}+{{\left( c+d \right)}^{2}}$ $\Rightarrow \sqrt{{{a}^{2}}+{{b}^{2}}}+\sqrt{{{c}^{2}}+{{d}^{2}}}\ge \sqrt{{{\left( a+b \right)}^{2}}+{{\left( c+d \right)}^{2}}}$
Hay $\left| {{w}_{1}}+{{w}_{2}} \right|\le \left| {{w}_{1}} \right|+\left| {{w}_{2}} \right|$ (*)
Giả sử ${{z}_{1}}={{a}_{1}}+{{b}_{1}}i; {{z}_{2}}={{c}_{1}}+{{d}_{1}}i;\ {{a}_{1}},{{b}_{1}},{{c}_{1}},{{d}_{1}}\in \mathbb{R}$, khi đó $a_{1}^{2}+b_{1}^{2}=1; c_{1}^{2}+d_{1}^{2}=7$.
Ta có $2={{\left| {{z}_{1}}-{{z}_{2}} \right|}^{2}}={{\left( {{a}_{1}}-{{c}_{1}} \right)}^{2}}+{{\left( {{b}_{1}}-{{d}_{1}} \right)}^{2}}=8-2\left( {{a}_{1}}{{c}_{1}}+{{b}_{1}}{{d}_{1}} \right)\Rightarrow {{a}_{1}}{{c}_{1}}+{{b}_{1}}{{d}_{1}}=3$
Mặt khác ${{\left| 3{{z}_{1}}+2{{z}_{2}} \right|}^{2}}={{\left| 3\left( {{a}_{1}}+{{b}_{1}}i \right)+2\left( {{c}_{1}}+{{d}_{1}}i \right) \right|}^{2}}={{\left( 3{{a}_{1}}+2{{c}_{1}} \right)}^{2}}+{{\left( 3{{b}_{1}}+2{{d}_{1}} \right)}^{2}}$
$=9\left( a_{1}^{2}+b_{1}^{2} \right)+4\left( c_{1}^{2}+d_{1}^{2} \right)+12\left( {{a}_{1}}{{c}_{1}}+{{b}_{1}}{{d}_{1}} \right)=73$.
Theo bất đẳng thức (*) ta có $\left| 3{{z}_{1}}+2{{z}_{2}}+{{z}_{3}} \right|\le \left| 3{{z}_{1}}+2{{z}_{2}} \right|+\left| {{z}_{3}} \right|=78\Rightarrow \left| {{z}_{3}} \right|=78-\sqrt{73}$.
Đáp án C.