Đáp án:
\(B.\ 12\)
Giải thích các bước giải:
\(\begin{array}{l}
\quad 9z^2 + 6z + 1 - m =0\\
\Leftrightarrow (3z + 1)^2 = m\qquad (*)\\
+)\quad m \geqslant 0\\
(*)\Leftrightarrow 3z + 1 = \pm \sqrt m\\
\Leftrightarrow 3z = -1 \pm \sqrt m\\
\Leftrightarrow z = \dfrac{-1 \pm\sqrt m}{3}\\
\Leftrightarrow |z| = \left|\dfrac{-1 \pm\sqrt m}{3}\right|\\
\Leftrightarrow 1 = \left|\dfrac{-1 \pm\sqrt m}{3}\right|\\
\Leftrightarrow \left[\begin{array}{l}\dfrac{-1 \pm\sqrt m}{3} = 1\\\dfrac{-1 \pm\sqrt m}{3}=-1\end{array}\right.\\
\Leftrightarrow \left[\begin{array}{l}-1 \pm \sqrt m = 3\\-1 \pm \sqrt m = -3\end{array}\right.\\
\Leftrightarrow \left[\begin{array}{l}\sqrt m = -4\\\sqrt m=4\\\sqrt m=2\\\sqrt m= -2\end{array}\right.\\
\Leftrightarrow \left[\begin{array}{l}m = 16\\m = 4\end{array}\right.\quad (Do\ m\in\Bbb R)\\
+)\quad m <0\\
(*) \Leftrightarrow 3z + 1 = \pm i\sqrt{-m}\\
\Leftrightarrow z = \dfrac{-1 \pm i\sqrt{-m}}{3}\\
\Leftrightarrow |z| = \left|\dfrac{-1 \pm i\sqrt{-m}}{3}\right|\\
\Leftrightarrow 1 = \sqrt{\left(-\dfrac13\right)^2 + \left(\dfrac{\sqrt{-m}}{3}\right)^2}\\
\Leftrightarrow 1 = \dfrac{1 - m}{9}\\
\Leftrightarrow 1 - m= 9\\
\Leftrightarrow m= -8\\
\text{Vậy}\ S = 16 + 4 - 8 = 12
\end{array}\)
