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Câu hỏi: Cho số phức $z=x+yi\left( x,y\in \mathbb{R} \right)$ thỏa mãn $\left( 1+2i \right)\overline{z}+z=3-4i$. Tính $S=x+4y$.
A. $4$.
B. $-3$.
C. $-4$.
D. $12$.
A. $4$.
B. $-3$.
C. $-4$.
D. $12$.
Ta có
$\begin{aligned}
& \left( 1+2i \right)\overline{z}+z=3-4i\Leftrightarrow \left( 1+2i \right)\left( x-yi \right)+x+yi=3-4i \\
& \Leftrightarrow \left\{ \begin{aligned}
& x+2y+x=3 \\
& 2x-y+y=-4 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& y=\dfrac{7}{2} \\
& x=-2 \\
\end{aligned} \right.\Rightarrow x+4y=12 \\
\end{aligned}$.
$\begin{aligned}
& \left( 1+2i \right)\overline{z}+z=3-4i\Leftrightarrow \left( 1+2i \right)\left( x-yi \right)+x+yi=3-4i \\
& \Leftrightarrow \left\{ \begin{aligned}
& x+2y+x=3 \\
& 2x-y+y=-4 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& y=\dfrac{7}{2} \\
& x=-2 \\
\end{aligned} \right.\Rightarrow x+4y=12 \\
\end{aligned}$.
Đáp án D.