Đáp án:
Giải thích các bước giải:
\(\begin{array}{l}
a.x - \sqrt {3{x^2} + 13x + 4} - 2 < 0\\
\to x - 2 < \sqrt {3{x^2} + 13x + 4} \\
\to \left\{ \begin{array}{l}
x > 2\\
{x^2} - 4x + 4 < 3{x^2} + 13x + 4
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x > 2\\
2{x^2} + 17x > 0
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x > 2\\
x \in \left( { - \infty ; - \frac{{17}}{2}} \right) \cup \left( {0; + \infty } \right)
\end{array} \right.\\
\to x > 2\\
b.2\sqrt {{x^2} - x - 12} \ge 4 - x\\
\to \left\{ \begin{array}{l}
x < 4\\
4\left( {{x^2} - x - 12} \right) \ge 16 - 8x + {x^2}
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x < 4\\
3{x^2} + 4x - 64 \ge 0
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x < 4\\
x \in \left( { - \infty ;\frac{{ - 16}}{3}} \right] \cup \left[ {4; + \infty } \right)
\end{array} \right.\\
\to x \in \left( { - \infty ;\frac{{ - 16}}{3}} \right]
\end{array}\)
