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Câu hỏi: Cho số phức $z=x+yi,(x,y\in \mathbb{R})$ thỏa mãn $(1+2i)\bar{z}+z=3-4i$. Tính giá trị biểu thức
$S=3x-2y.~$
A. $S=-12.~~~~~~~~$
B. $S=-11.~~~~~~~~$
C. $S=-13.~~~~~~~~$
D. $S=-10.~$
$S=3x-2y.~$
A. $S=-12.~~~~~~~~$
B. $S=-11.~~~~~~~~$
C. $S=-13.~~~~~~~~$
D. $S=-10.~$
Cách giải:
Ta có :
$\begin{aligned}
& (1+2i)\overline{z}+z=3-4i \\
& \Leftrightarrow (1+2i)(x-yi)+x+yi=3-4i \\
\end{aligned}$
$\begin{aligned}
& \Leftrightarrow x-yi+2xi+2y+x+yi=3-4i \\
& \Leftrightarrow 2x+2y+2xi=3-4i \\
\end{aligned}$
$\Leftrightarrow \left\{ \begin{array}{*{35}{l}}
2x+2y=3 \\
2x=-4 \\
\end{array}\Leftrightarrow \left\{ \begin{array}{*{35}{l}}
y=\dfrac{7}{2} \\
x=-2 \\
\end{array} \right. \right.$
$S=3x-2y=3.(-2)-2.\dfrac{7}{2}=-6-7=-13$
Ta có :
$\begin{aligned}
& (1+2i)\overline{z}+z=3-4i \\
& \Leftrightarrow (1+2i)(x-yi)+x+yi=3-4i \\
\end{aligned}$
$\begin{aligned}
& \Leftrightarrow x-yi+2xi+2y+x+yi=3-4i \\
& \Leftrightarrow 2x+2y+2xi=3-4i \\
\end{aligned}$
$\Leftrightarrow \left\{ \begin{array}{*{35}{l}}
2x+2y=3 \\
2x=-4 \\
\end{array}\Leftrightarrow \left\{ \begin{array}{*{35}{l}}
y=\dfrac{7}{2} \\
x=-2 \\
\end{array} \right. \right.$
$S=3x-2y=3.(-2)-2.\dfrac{7}{2}=-6-7=-13$
Đáp án C.