\[\begin{array}{l}
a){\rm{ }}{x^2}{\rm{ }} - {\rm{ }}3x{\rm{ }} - {\rm{ }}4 \le 0\\
 \Leftrightarrow {x^2} + x - 4x - 4 \le 0\\
 \Leftrightarrow x\left( {x + 1} \right) - 4\left( {x + 1} \right) \le 0\\
 \Leftrightarrow \left( {x + 1} \right)\left( {x - 4} \right) \le 0\\
 \Leftrightarrow \left[ {\begin{array}{*{20}{c}}
{\left\{ \begin{array}{l}
x + 1 \le 0\\
x - 4 \ge 0
\end{array} \right.}\\
{\left\{ \begin{array}{l}
x + 1 \ge 0\\
x - 4 \le 0
\end{array} \right.}
\end{array}} \right.\\
 \Leftrightarrow \left[ {\begin{array}{*{20}{c}}
{\left\{ \begin{array}{l}
x \le  - 1\\
x \ge 4
\end{array} \right.}\\
{\left\{ \begin{array}{l}
x \ge  - 1\\
x \le 4
\end{array} \right.}
\end{array}} \right.\\
 \Leftrightarrow  - 1 \le x \le 4\\
vay: - 1 \le x \le 4\\
b){x^2}{\rm{ }} - {\rm{ }}x{\rm{ }} - {\rm{ }}6 \ge 0\\
 =  > {x^2} - 3x + 2x - 6 \ge 0\\
 =  > x\left( {x - 3} \right) + 2\left( {x - 3} \right) \ge 0\\
 =  > \left( {x - 3} \right)\left( {x + 2} \right) \ge 0\\
 =  > \left[ {\begin{array}{*{20}{c}}
{\left\{ \begin{array}{l}
x - 3 \ge 0\\
x + 2 \ge 0
\end{array} \right.}\\
{\left\{ \begin{array}{l}
x - 3 \le 0\\
x + 2 \le 0
\end{array} \right.}
\end{array}} \right.\\
 =  > \left[ {\begin{array}{*{20}{c}}
{\left\{ \begin{array}{l}
x \ge 3\\
x \ge  - 2
\end{array} \right.}\\
{\left\{ \begin{array}{l}
x \le 3\\
x \le  - 2
\end{array} \right.}
\end{array}} \right.\\
 =  > \left[ \begin{array}{l}
x \ge 3\\
x \le  - 2
\end{array} \right.\\
vay:\left[ \begin{array}{l}
x \ge 3\\
x \le  - 2
\end{array} \right.\\
c){\rm{ }} - {x^2} + {\rm{ }}2x{\rm{ }} - {\rm{ }}6 > 0\\
 =  > {x^2} - 2x + 6 < 0\\
 =  > {x^2} - 2x + 1 + 5 < 0\\
 =  > {\left( {x - 1} \right)^2} + 5 < 0\\
do:{\left( {x - 1} \right)^2} \ge 0\\
 =  > {\left( {x - 1} \right)^2} + 5 \ge 5 > 0
\end{array}\]

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