Đáp án:
\(x \in \left( { - 1;1} \right] \cup \left[ {2;4} \right)\)
Giải thích các bước giải:
\(\begin{array}{l}
\sqrt {6{x^2} - 18x + 12} < 3x + 10 - {x^2}\\
\to \sqrt {6\left( {{x^2} - 3x + 2} \right)} + {x^2} - 3x - 10 < 0\left( 1 \right)\\
Đặt:\sqrt {6\left( {{x^2} - 3x + 2} \right)} = t\left( {t \ge 0} \right)\\
\to {t^2} = 6\left( {{x^2} - 3x + 2} \right)\\
\to {x^2} - 3x + 2 = \dfrac{{{t^2}}}{6}\\
\left( 1 \right) \to t + \dfrac{{{t^2}}}{6} - 12 < 0\\
\to {t^2} + 6t - 72 < 0\\
\to \left( {t - 6} \right)\left( {t + 12} \right) < 0\\
\to t \in \left( { - 12;6} \right)\\
\to t \in \left[ {0;6} \right)\\
\to \left\{ \begin{array}{l}
\sqrt {6\left( {{x^2} - 3x + 2} \right)} \ge 0\\
\sqrt {6\left( {{x^2} - 3x + 2} \right)} < 6
\end{array} \right.\\
\to \left\{ \begin{array}{l}
{x^2} - 3x + 2 \ge 0\\
6\left( {{x^2} - 3x + 2} \right) < 36
\end{array} \right.\\
\to \left\{ \begin{array}{l}
\left( {x - 2} \right)\left( {x - 1} \right) \ge 0\\
{x^2} - 3x + 2 - 6 < 0
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x \in \left( { - \infty ;1} \right] \cup \left[ {2; + \infty } \right)\\
\left( {x - 4} \right)\left( {x + 1} \right) < 0
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x \in \left( { - \infty ;1} \right] \cup \left[ {2; + \infty } \right)\\
x \in \left( { - 1;4} \right)
\end{array} \right.\\
KL:x \in \left( { - 1;1} \right] \cup \left[ {2;4} \right)
\end{array}\)
