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Câu hỏi: Cho hai số phức z, w thỏa mãn $\left| z+2w \right|=3,\left| 2\text{z}+3w \right|=6$ và $\left| z+4w \right|=7$. Tính giá trị của biểu thức $P=z.\overline{\text{w}}+\overline{z}.\text{w}$.
A. $P=-14i$
B. $P=-28i$
C. $P=-14$
D. $P=-28$
A. $P=-14i$
B. $P=-28i$
C. $P=-14$
D. $P=-28$
Ta có: $\left| z+2w \right|=3\Leftrightarrow {{\left| z+2\text{w} \right|}^{2}}=9\Leftrightarrow \left( z+2w \right).\left( \overline{z+2w} \right)=9\Leftrightarrow \left( z+2w \right)\left( \overline{z}+2\overline{\text{w}} \right)=9$
$\Leftrightarrow z.\overline{z}+2\left( z.\overline{\text{w}}+\overline{z}.\text{w} \right)+4w.\overline{\text{w}}=9\Leftrightarrow {{\left| z \right|}^{2}}+2P+4{{\left| \text{w} \right|}^{2}}=9$ (1).
Tương tự:
$\left| 2\text{z}+3w \right|=6\Leftrightarrow {{\left| 2\text{z}+3w \right|}^{2}}=36\Leftrightarrow \left( 2\text{z}+3w \right)\left( 2\overline{z}+3\overline{\text{w}} \right)=36\Leftrightarrow 4{{\left| z \right|}^{2}}+6P+9{{\left| \text{w} \right|}^{2}}=36$ (2).
$\left| z+4w \right|=7\Leftrightarrow \left( z+4w \right)\left( \overline{z}+4\overline{\text{w}} \right)=49\Leftrightarrow {{\left| z \right|}^{2}}+4P+16{{\left| \text{w} \right|}^{2}}=49$ (3).
Giải hệ phương trình gồm (1), (2), (3) ta có: $\left\{ \begin{aligned}
& {{\left| z \right|}^{2}}=33 \\
& P=-28 \\
& {{\left| \text{w} \right|}^{2}}=8 \\
\end{aligned} \right.\Rightarrow P=-28$.
$\Leftrightarrow z.\overline{z}+2\left( z.\overline{\text{w}}+\overline{z}.\text{w} \right)+4w.\overline{\text{w}}=9\Leftrightarrow {{\left| z \right|}^{2}}+2P+4{{\left| \text{w} \right|}^{2}}=9$ (1).
Tương tự:
$\left| 2\text{z}+3w \right|=6\Leftrightarrow {{\left| 2\text{z}+3w \right|}^{2}}=36\Leftrightarrow \left( 2\text{z}+3w \right)\left( 2\overline{z}+3\overline{\text{w}} \right)=36\Leftrightarrow 4{{\left| z \right|}^{2}}+6P+9{{\left| \text{w} \right|}^{2}}=36$ (2).
$\left| z+4w \right|=7\Leftrightarrow \left( z+4w \right)\left( \overline{z}+4\overline{\text{w}} \right)=49\Leftrightarrow {{\left| z \right|}^{2}}+4P+16{{\left| \text{w} \right|}^{2}}=49$ (3).
Giải hệ phương trình gồm (1), (2), (3) ta có: $\left\{ \begin{aligned}
& {{\left| z \right|}^{2}}=33 \\
& P=-28 \\
& {{\left| \text{w} \right|}^{2}}=8 \\
\end{aligned} \right.\Rightarrow P=-28$.
Đáp án D.