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Câu hỏi: Môđun của số phức $z$ thỏa mãn $\left( 2-i \right)z+\left( 1-i \right)\overline{z}=9-8i$ bằng
A. $1$.
B. $\sqrt{5}$.
C. $\sqrt{13}$.
D. $5$.
A. $1$.
B. $\sqrt{5}$.
C. $\sqrt{13}$.
D. $5$.
Gọi $z=a+bi \left( a,b\in \mathbb{R} \right)\Rightarrow \overline{z}=a-bi$, ta có:
$\begin{aligned}
& \left( 2-i \right)\left( a+bi \right)+\left( 1-i \right)\left( a-bi \right)=9-8i \\
& \Leftrightarrow 2a+b+\left( 2b-a \right)i+a-b-\left( a+b \right)i=9-8i \\
& \Leftrightarrow \left\{ \begin{aligned}
& 2a+b+a-b=9 \\
& 2b-a-a-b=-8 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& a=3 \\
& b=2a-8 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& a=3 \\
& b=-2 \\
\end{aligned} \right. \\
& \left| z \right|=\sqrt{{{3}^{2}}+{{\left( -2 \right)}^{2}}}=\sqrt{13}. \\
\end{aligned}$
$\begin{aligned}
& \left( 2-i \right)\left( a+bi \right)+\left( 1-i \right)\left( a-bi \right)=9-8i \\
& \Leftrightarrow 2a+b+\left( 2b-a \right)i+a-b-\left( a+b \right)i=9-8i \\
& \Leftrightarrow \left\{ \begin{aligned}
& 2a+b+a-b=9 \\
& 2b-a-a-b=-8 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& a=3 \\
& b=2a-8 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& a=3 \\
& b=-2 \\
\end{aligned} \right. \\
& \left| z \right|=\sqrt{{{3}^{2}}+{{\left( -2 \right)}^{2}}}=\sqrt{13}. \\
\end{aligned}$
Đáp án C.