Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
2,\\
\left| {{x^2} + x - 6} \right| = 4x \Leftrightarrow \left\{ \begin{array}{l}
4x \ge 0\\
\left[ \begin{array}{l}
{x^2} + x - 6 = 4x\\
{x^2} + x - 6 = - 4x
\end{array} \right.
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
x \ge 0\\
\left[ \begin{array}{l}
{x^2} - 3x - 6 = 0\\
{x^2} + 5x - 6 = 0
\end{array} \right.
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x \ge 0\\
\left[ \begin{array}{l}
x = \frac{{3 \pm \sqrt {33} }}{2}\\
x = 1\\
x = - 6
\end{array} \right.
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = \frac{{3 + \sqrt {33} }}{2}\\
x = 1
\end{array} \right.\\
3,\\
\left| {{x^2} - 3x} \right| + \left| {x - 1} \right| = 2\,\,\,\,\,\,\,\,\,\,\,\left( 1 \right)\\
TH1:\,\,\,x < 1 \Rightarrow x - 1 < 0 \Rightarrow \left| {x - 1} \right| = 1 - x\\
\left( 1 \right) \Leftrightarrow \left| {{x^2} - 3x} \right| + 1 - x = 2\\
\Leftrightarrow \left| {{x^2} - 3x} \right| = x + 1\\
\Leftrightarrow \left\{ \begin{array}{l}
x + 1 \ge 0\\
\left[ \begin{array}{l}
{x^2} - 3x = x + 1\\
{x^2} - 3x = - x - 1
\end{array} \right.
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x \ge - 1\\
\left[ \begin{array}{l}
{x^2} - 4x - 1 = 0\\
{x^2} - 2x + 1 = 0
\end{array} \right.
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
x \ge - 1\\
\left[ \begin{array}{l}
x = 2 \pm \sqrt 5 \\
x = 1
\end{array} \right.
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 2 \pm \sqrt 5 \\
x = 1
\end{array} \right.\\
x < 1 \Rightarrow x = 2 - \sqrt 5 \\
TH2:\,\,x \ge 1 \Leftrightarrow x - 1 \ge 0 \Rightarrow \left| {x - 1} \right| = x - 1\\
\left( 1 \right) \Leftrightarrow \left| {{x^2} - 3x} \right| + x - 1 = 2\\
\Leftrightarrow \left| {{x^2} - 3x} \right| = 3 - x\\
\Leftrightarrow \left\{ \begin{array}{l}
3 - x \ge 0\\
\left[ \begin{array}{l}
{x^2} - 3x = 3 - x\\
{x^2} - 3x = x - 3
\end{array} \right.
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x \le 3\\
\left[ \begin{array}{l}
{x^2} - 2x - 3 = 0\\
{x^2} - 4x + 3 = 0
\end{array} \right.
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
x \le 3\\
\left[ \begin{array}{l}
x = 3\\
x = - 1\\
x = 1
\end{array} \right.
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 3\\
x = - 1\\
x = 1
\end{array} \right.\\
x \ge 1 \Rightarrow \left[ \begin{array}{l}
x = 1\\
x = 3
\end{array} \right.
\end{array}\)
