Đáp án:
b) \(\left[ \begin{array}{l}
x = \dfrac{1}{2}\\
x = \dfrac{3}{2}
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\left| {{x^2} - x} \right| = - 2x\\
\to \left[ \begin{array}{l}
{x^2} - x = - 2x\left( {x \le 0} \right)\\
{x^2} - x = 2x\left( {x > 0} \right)
\end{array} \right.\\
\to \left[ \begin{array}{l}
{x^2} + x = 0\\
{x^2} - 3x = 0
\end{array} \right.\\
\to \left[ \begin{array}{l}
x\left( {x + 1} \right) = 0\\
x\left( {x - 3} \right) = 0
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 0\\
x = - 1\\
x = 3
\end{array} \right.\\
b)12{x^2} - 24x + 9 = 0\\
\to 12{x^2} - 6x - 18x + 9 = 0\\
\to 6x\left( {2x - 1} \right) - 9\left( {2x - 1} \right) = 0\\
\to \left( {2x - 1} \right)\left( {6x - 9} \right) = 0\\
\to \left[ \begin{array}{l}
x = \dfrac{1}{2}\\
x = \dfrac{3}{2}
\end{array} \right.
\end{array}\)
