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Câu hỏi: Cho số phức $z$ thỏa mãn $\left( 2+3i \right)z=z-1$. Môđun của $\overline{z}$ bằng
A. $\dfrac{1}{\sqrt{10}}$.
B. $\dfrac{1}{10}$.
C. $1$.
D. $\sqrt{10}$.
A. $\dfrac{1}{\sqrt{10}}$.
B. $\dfrac{1}{10}$.
C. $1$.
D. $\sqrt{10}$.
Ta có $\left( 2+3i \right)z=z-1$
$\Leftrightarrow \left( 1+3i \right)z=-1$
$\Leftrightarrow z=\dfrac{-1}{1+3i}$
$\Leftrightarrow z=\dfrac{-1.\left( 1-3i \right)}{10}$
$\Leftrightarrow z=\dfrac{-1}{10}+\dfrac{3i}{10}$
$\Rightarrow \overline{z}=\dfrac{-1}{10}-\dfrac{3i}{10}$.
Vậy $\left| \overline{z} \right|=\sqrt{{{\left( \dfrac{-1}{10} \right)}^{2}}+{{\left( \dfrac{-3}{10} \right)}^{2}}}=\dfrac{1}{\sqrt{10}}$.
$\Leftrightarrow \left( 1+3i \right)z=-1$
$\Leftrightarrow z=\dfrac{-1}{1+3i}$
$\Leftrightarrow z=\dfrac{-1.\left( 1-3i \right)}{10}$
$\Leftrightarrow z=\dfrac{-1}{10}+\dfrac{3i}{10}$
$\Rightarrow \overline{z}=\dfrac{-1}{10}-\dfrac{3i}{10}$.
Vậy $\left| \overline{z} \right|=\sqrt{{{\left( \dfrac{-1}{10} \right)}^{2}}+{{\left( \dfrac{-3}{10} \right)}^{2}}}=\dfrac{1}{\sqrt{10}}$.
Đáp án A.
