Đáp án:
$\begin{array}{l}
a)\left( {x + 1} \right)\left( {x - 1} \right)\left( {x - 2} \right) > 0\\
\Rightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x + 1 > 0\\
\left( {x - 1} \right)\left( {x - 2} \right) > 0
\end{array} \right.\\
\left\{ \begin{array}{l}
x + 1 < 0\\
\left( {x - 1} \right)\left( {x - 2} \right) < 0
\end{array} \right.
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x > - 1\\
\left[ \begin{array}{l}
x > 2\\
x < 1
\end{array} \right.
\end{array} \right.\\
\left\{ \begin{array}{l}
x < - 1\\
1 < x < 2
\end{array} \right.\left( {ktm} \right)
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
x > 2\\
- 1 < x < 1
\end{array} \right.\\
Vậy\,x > 2\,hoặc\, - 1 < x < 1\\
b)\left( {2x - 7} \right)\left( {5 - x} \right) \ge 0\\
\Rightarrow \left( {2x - 7} \right)\left( {x - 5} \right) \le 0\\
\Rightarrow \frac{7}{2} \le x \le 5\\
Vậy\,\frac{7}{2} \le x \le 5\\
c){x^2} - x - 20 - 2\left( {x - 11} \right) > 0\\
\Rightarrow {x^2} - x - 20 - 2x + 22 > 0\\
\Rightarrow {x^2} - 3x + 2 > 0\\
\Rightarrow \left( {x - 1} \right)\left( {x - 2} \right) > 0\\
\Rightarrow \left[ \begin{array}{l}
x > 2\\
x < 1
\end{array} \right.\\
Vậy\,x < 1\,hoặc\,x > 2\\
d){x^3} + 8{x^2} + 17x + 10 < 0\\
\Rightarrow {x^3} + 5{x^2} + 3{x^2} + 15x + 2x + 10 < 0\\
\Rightarrow \left( {x + 5} \right)\left( {{x^2} + 3x + 2} \right) < 0\\
\Rightarrow \left( {x + 5} \right)\left( {x + 2} \right)\left( {x + 1} \right) < 0\\
\Rightarrow \left[ \begin{array}{l}
x < - 5\\
- 2 < x < - 1
\end{array} \right.\\
Vậy\,x < - 5\,hoặc\, - 2 < x < - 1
\end{array}$
