Google
Câu hỏi: Cho số phức $z$ thỏa $2z+3\bar{z}=10+i$. Tính $\left| z \right|$.
A. $\left| z \right|=5$.
B. $\left| z \right|=3$.
C. $\left| z \right|=\sqrt{3}$.
D. $\left| z \right|=\sqrt{5}$.
A. $\left| z \right|=5$.
B. $\left| z \right|=3$.
C. $\left| z \right|=\sqrt{3}$.
D. $\left| z \right|=\sqrt{5}$.
Gọi $z=a+bi\Rightarrow \bar{z}=a-bi$, $\left( a,b\in \mathbb{R} \right)$.
Ta có: $2\left( a+bi \right)+3(a-bi)=10+i\Leftrightarrow \left\{ \begin{aligned}
& 5a=10 \\
& -b=1 \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& a=2 \\
& b=-1 \\
\end{aligned} \right.\Rightarrow z=2-i$.
Vậy $\left| z \right|=\sqrt{{{2}^{2}}+{{\left( -1 \right)}^{2}}}=\sqrt{5}$.
Ta có: $2\left( a+bi \right)+3(a-bi)=10+i\Leftrightarrow \left\{ \begin{aligned}
& 5a=10 \\
& -b=1 \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& a=2 \\
& b=-1 \\
\end{aligned} \right.\Rightarrow z=2-i$.
Vậy $\left| z \right|=\sqrt{{{2}^{2}}+{{\left( -1 \right)}^{2}}}=\sqrt{5}$.
Đáp án D.