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Câu hỏi: Biết $z=a+bi,\left( a,b\in \mathbb{R} \right)$ là số phức thỏa mãn $\left( 3-2i \right)z-2i\overline{z}=15-8i$. Tổng $2a+b$ là
A. $2a+b=5$.
B. $2a+b=14$.
C. $2a+b=9$.
D. $2a+b=12$.
A. $2a+b=5$.
B. $2a+b=14$.
C. $2a+b=9$.
D. $2a+b=12$.
Ta có
$\begin{aligned}
& \left( 3-2i \right)z-2i\overline{z}=15-8i \\
& \Leftrightarrow \left( 3-2i \right)\left( a+bi \right)-2i\left( a-bi \right)=15-8i \\
& \Leftrightarrow 3a+2b+3bi-2ai-2ai-2b=15-8i \\
& \Leftrightarrow 3a+\left( 3b-4a \right)i=15-8i \\
& \Leftrightarrow \left\{ \begin{aligned}
& 3a=15 \\
& 3b-4a=-8 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& a=5 \\
& b=4 \\
\end{aligned} \right.\Rightarrow 2a+b=14 \\
\end{aligned}$.
$\begin{aligned}
& \left( 3-2i \right)z-2i\overline{z}=15-8i \\
& \Leftrightarrow \left( 3-2i \right)\left( a+bi \right)-2i\left( a-bi \right)=15-8i \\
& \Leftrightarrow 3a+2b+3bi-2ai-2ai-2b=15-8i \\
& \Leftrightarrow 3a+\left( 3b-4a \right)i=15-8i \\
& \Leftrightarrow \left\{ \begin{aligned}
& 3a=15 \\
& 3b-4a=-8 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& a=5 \\
& b=4 \\
\end{aligned} \right.\Rightarrow 2a+b=14 \\
\end{aligned}$.
Đáp án B.