Cho các số phức z,w khác 0 thỏa mãn z + w $\neq$ 0 và $\frac{1}{z}$+ $\frac{3}{w}$ = $\frac{6}{z+w}$ Khi đó môđun của $\frac{z}{w}$ bằng A: $\sqrt{3}$ B: $\frac{1}{\sqrt{3} }$ C: 3 D: $\frac{1}{3}$

Đáp án:

$B.\ \dfrac{1}{\sqrt3}$

Giải thích các bước giải:

$\quad \dfrac1z +\dfrac3w =\dfrac{6}{z+w}$

$\Leftrightarrow \dfrac{w + 3z}{zw}=\dfrac{6}{z+w}$

$\Leftrightarrow (w+3z)(z+w) = 6zw$

$\Leftrightarrow w^2 + 4zw + 3z^2 = 6zw$

$\Leftrightarrow w^2 - 2zw + z^2 = - 2z^2$

$\Leftrightarrow (w - z)^2 = \left(\sqrt2iz\right)^2$

$\Leftrightarrow \left[\begin{array}{l}w - z = \sqrt2iz\\w -z= -\sqrt2iz\end{array}\right.$

$\Leftrightarrow \left[\begin{array}{l}w= z + \sqrt2iz\\w =z -\sqrt2iz\end{array}\right.$

$\Leftrightarrow \left[\begin{array}{l}w= z\left(1 + \sqrt2i\right)\\w =z\left(1-\sqrt2i\right)\end{array}\right.$

$\Leftrightarrow \left[\begin{array}{l}\dfrac{z}{w}=\dfrac{1}{1+\sqrt2i}\\\dfrac{z}{w}=\dfrac{1}{1 - \sqrt2i}\end{array}\right.$

$\Leftrightarrow \left[\begin{array}{l}\left|\dfrac{z}{w}\right|=\left|\dfrac{1}{1+\sqrt2i}\right|\\\left|\dfrac{z}{w}\right|=\left|\dfrac{1}{1 - \sqrt2i}\right|\end{array}\right.$

$\Leftrightarrow \left[\begin{array}{l}\left|\dfrac{z}{w}\right|=\dfrac{1}{\sqrt{1^2 + \left(\sqrt2\right)^2}}\\\left|\dfrac{z}{w}\right|=\dfrac{1}{\sqrt{1^2 +\left(- \sqrt2\right)^2}}\end{array}\right.$

$\Leftrightarrow \left|\dfrac{z}{w}\right|=\dfrac{1}{\sqrt3}$