Đáp án:
$\begin{array}{l}
a)\sqrt {\left( {x + 5} \right)\left( {3x + 4} \right)} > 4\left( {x - 1} \right)\\
\Rightarrow 3{x^2} + 19x + 20 > 16\left( {{x^2} - 2x + 1} \right)\\
\Rightarrow 13{x^2} - 51x - 4 < 0\\
\Rightarrow \left( {13x + 1} \right)\left( {x - 4} \right) < 0\\
\Rightarrow - \frac{1}{{13}} < x < 4\\
b)\left| { - {x^2} + 3x + 2} \right| < {x^2} - 3x + 2\\
\Rightarrow \left\{ \begin{array}{l}
{x^2} - 3x + 2 > 0\\
\left[ \begin{array}{l}
- {x^2} + 3x + 2 < {x^2} - 3x + 2\\
- {x^2} + 3x + 2 > - {x^2} + 3x - 2
\end{array} \right.
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
x > 2/x < 1\\
\left[ \begin{array}{l}
2{x^2} - 6x > 0\\
2 > - 2\left( {ld} \right)
\end{array} \right.
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
x > 2/x < 1\\
x\left( {x - 3} \right) > 0
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
x < 0\\
x > 3
\end{array} \right.
\end{array}$
