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Câu hỏi: Cho số phức $z=a+bi\left( a;b\in \mathbb{R} \right)$ thỏa mãn $\left( 1+2i \right)z-\left( 2-3i \right)\overline{z}=2+30i.$ Tổng $a+b$ có giá trị bằng
A. $-8.$
B. $-2.$
C. $2.$
D. $8.$
A. $-8.$
B. $-2.$
C. $2.$
D. $8.$
Ta có $z=a+bi\Rightarrow \overline{z}=a-bi$
Khi đó $\left( 1+2i \right)z-\left( 2-3i \right)\overline{z}=2+30i$
$\Leftrightarrow \left( 1+2i \right)\left( a+bi \right)-\left( 2-3i \right)\left( a-bi \right)=2+30i$
$\Leftrightarrow a+bi+2ai-2b-2a+2bi+3ai+3b=2+30i$
$\Leftrightarrow -a+b+\left( 5a+3b \right)i=2+30i$
$\Leftrightarrow \left\{ \begin{aligned}
& -a+b=2 \\
& 5a+3b=30 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& a=3 \\
& b=5 \\
\end{aligned} \right.\Rightarrow a+b=8.$
Khi đó $\left( 1+2i \right)z-\left( 2-3i \right)\overline{z}=2+30i$
$\Leftrightarrow \left( 1+2i \right)\left( a+bi \right)-\left( 2-3i \right)\left( a-bi \right)=2+30i$
$\Leftrightarrow a+bi+2ai-2b-2a+2bi+3ai+3b=2+30i$
$\Leftrightarrow -a+b+\left( 5a+3b \right)i=2+30i$
$\Leftrightarrow \left\{ \begin{aligned}
& -a+b=2 \\
& 5a+3b=30 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& a=3 \\
& b=5 \\
\end{aligned} \right.\Rightarrow a+b=8.$
Đáp án D.