Đáp án:
$B.\ \dfrac{1}{\sqrt3}$
Giải thích các bước giải:
$\quad \dfrac1z +\dfrac3w =\dfrac{6}{z+w}$
$\Leftrightarrow \dfrac{w + 3z}{zw}=\dfrac{6}{z+w}$
$\Leftrightarrow (w+3z)(z+w) = 6zw$
$\Leftrightarrow w^2 + 4zw + 3z^2 = 6zw$
$\Leftrightarrow w^2 - 2zw + z^2 = - 2z^2$
$\Leftrightarrow (w - z)^2 = \left(\sqrt2iz\right)^2$
$\Leftrightarrow \left[\begin{array}{l}w - z = \sqrt2iz\\w -z= -\sqrt2iz\end{array}\right.$
$\Leftrightarrow \left[\begin{array}{l}w= z + \sqrt2iz\\w =z -\sqrt2iz\end{array}\right.$
$\Leftrightarrow \left[\begin{array}{l}w= z\left(1 + \sqrt2i\right)\\w =z\left(1-\sqrt2i\right)\end{array}\right.$
$\Leftrightarrow \left[\begin{array}{l}\dfrac{z}{w}=\dfrac{1}{1+\sqrt2i}\\\dfrac{z}{w}=\dfrac{1}{1 - \sqrt2i}\end{array}\right.$
$\Leftrightarrow \left[\begin{array}{l}\left|\dfrac{z}{w}\right|=\left|\dfrac{1}{1+\sqrt2i}\right|\\\left|\dfrac{z}{w}\right|=\left|\dfrac{1}{1 - \sqrt2i}\right|\end{array}\right.$
$\Leftrightarrow \left[\begin{array}{l}\left|\dfrac{z}{w}\right|=\dfrac{1}{\sqrt{1^2 + \left(\sqrt2\right)^2}}\\\left|\dfrac{z}{w}\right|=\dfrac{1}{\sqrt{1^2 +\left(- \sqrt2\right)^2}}\end{array}\right.$
$\Leftrightarrow \left|\dfrac{z}{w}\right|=\dfrac{1}{\sqrt3}$
