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Câu hỏi: Cho số phức $z=a+bi \left( a,b\in \mathbb{R} \right)$ thỏa mãn $3z-\left( 4+5i \right)\overline{z}=-17+11i.$ Tính $ab.$
A. $ab=-3.$
B. $ab=3.$
C. $ab=6.$
D. $ab=-6.$
A. $ab=-3.$
B. $ab=3.$
C. $ab=6.$
D. $ab=-6.$
Theo bài ra ta có $3z-\left( 4+5i \right)\overline{z}=-17+11i\Leftrightarrow 3\left( a+bi \right)-\left( 4+5i \right)\left( a-bi \right)=-17+11i$
$\Leftrightarrow 3a+3bi-\left( 4a-4bi+5ai+5b \right)=-17+11i$
$\Leftrightarrow 3a+3bi-4a+4bi-5ai-5b=-17+11i$
$\Leftrightarrow -a-5b+7bi-5ai=-17+11i$
$\Leftrightarrow \left( -a-5b \right)+\left( -5a+7b \right)i=-17+11i$ $\Rightarrow \left\{ \begin{aligned}
& -a-5b=-17 \\
& -5a+7b=11 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& a=2 \\
& b=3 \\
\end{aligned} \right.$
Do đó $ab=6.$
$\Leftrightarrow 3a+3bi-\left( 4a-4bi+5ai+5b \right)=-17+11i$
$\Leftrightarrow 3a+3bi-4a+4bi-5ai-5b=-17+11i$
$\Leftrightarrow -a-5b+7bi-5ai=-17+11i$
$\Leftrightarrow \left( -a-5b \right)+\left( -5a+7b \right)i=-17+11i$ $\Rightarrow \left\{ \begin{aligned}
& -a-5b=-17 \\
& -5a+7b=11 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& a=2 \\
& b=3 \\
\end{aligned} \right.$
Do đó $ab=6.$
Đáp án C.