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Câu hỏi: Cho số phức ${z=x+yi}$ ( với ${x,y\in \mathbb{R}}$ ) thỏa mãn ${z-\left( 2+3i \right)\overline{z}=1-9i}$. Tính ${xy}$.
A. $xy=-1$.
B. $xy=1$.
C. $xy=-2$.
D. $xy=2$.
A. $xy=-1$.
B. $xy=1$.
C. $xy=-2$.
D. $xy=2$.
Ta có:
$\begin{aligned}
& z-\left( 2+3i \right)\overline{z}=1-9i \\
& \Leftrightarrow x+yi-\left( 2+3i \right)\left( x-yi \right)=1-9i \\
& \Leftrightarrow \left( -x-3y \right)+\left( -3x+3y \right)i=1-9i \\
& \Leftrightarrow \left\{ \begin{aligned}
& -x-3y=1 \\
& -3x+3y=-9 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& x=2 \\
& y=-1 \\
\end{aligned} \right. \\
\end{aligned}$:
$\Rightarrow xy=-2$.
$\begin{aligned}
& z-\left( 2+3i \right)\overline{z}=1-9i \\
& \Leftrightarrow x+yi-\left( 2+3i \right)\left( x-yi \right)=1-9i \\
& \Leftrightarrow \left( -x-3y \right)+\left( -3x+3y \right)i=1-9i \\
& \Leftrightarrow \left\{ \begin{aligned}
& -x-3y=1 \\
& -3x+3y=-9 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& x=2 \\
& y=-1 \\
\end{aligned} \right. \\
\end{aligned}$:
$\Rightarrow xy=-2$.
Đáp án C.
