Đáp án:
\[\left[ \begin{array}{l}
x = - \dfrac{1}{{12}}\\
x = \dfrac{1}{6}
\end{array} \right.\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
- 16{x^2} + \dfrac{4}{3}x + \dfrac{2}{9} = 0\\
\Leftrightarrow - 144{x^2} + 12x + 2 = 0\\
\Leftrightarrow 144{x^2} - 12x - 2 = 0\\
\Leftrightarrow \left( {144{x^2} + 12x} \right) - \left( {24x + 2} \right) = 0\\
\Leftrightarrow 12x.\left( {12x + 1} \right) - 2.\left( {12x + 1} \right) = 0\\
\Leftrightarrow \left( {12x + 1} \right)\left( {12x - 2} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
12x + 1 = 0\\
12x - 2 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = - \dfrac{1}{{12}}\\
x = \dfrac{1}{6}
\end{array} \right.
\end{array}\)
