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Câu hỏi: Cho các số phức $z, w$ khác 0 thỏa mãn $z+w \neq 0$ và $\dfrac{1}{z}+\dfrac{3}{w}=\dfrac{6}{z+w}$. Khi đó $\left|\dfrac{z}{w}\right|$ bằng
A. $\dfrac{1}{\sqrt{3}}$.
B. $\dfrac{1}{3}$.
C. $\sqrt{3}$.
D. 3 .
A. $\dfrac{1}{\sqrt{3}}$.
B. $\dfrac{1}{3}$.
C. $\sqrt{3}$.
D. 3 .
Ta có: $\dfrac{1}{z}+\dfrac{3}{w}=\dfrac{6}{z+w} \Leftrightarrow \dfrac{w+3 z}{z w}=\dfrac{6}{z+w} \Leftrightarrow(w+3 z)(z+w)=6 z w \Leftrightarrow 3 z^2-2 z w+w^2=0$ $\Leftrightarrow 3\left(\dfrac{z}{w}\right)^2-2 \dfrac{z}{w}+1=0 \Leftrightarrow \dfrac{z}{w}=\dfrac{1}{3} \pm \dfrac{\sqrt{2}}{3} i$.
Vậy, $\left|\dfrac{\underline{z}}{w}\right|=\sqrt{\left(\dfrac{1}{3}\right)^2+\left(\dfrac{\sqrt{2}}{3}\right)^2}=\dfrac{1}{\sqrt{3}}$.
Vậy, $\left|\dfrac{\underline{z}}{w}\right|=\sqrt{\left(\dfrac{1}{3}\right)^2+\left(\dfrac{\sqrt{2}}{3}\right)^2}=\dfrac{1}{\sqrt{3}}$.
Đáp án A.