Đáp án:
$\begin{array}{l}
1)a) - 4{x^2} + x - 3 > 0\\
\Rightarrow 4{x^2} - x + 3 > 0\\
\Rightarrow 4{x^2} - 2.2x.\dfrac{1}{4} + \dfrac{1}{{16}} + \dfrac{{47}}{{16}} > 0\\
\Rightarrow {\left( {2x - \dfrac{1}{4}} \right)^2} + \dfrac{{47}}{{16}} > 0\left( {tm} \right)\\
Vậy\,x \in R\\
b)3{x^2} + x + 5 > 0\\
\Rightarrow {x^2} + \dfrac{1}{3}.x + \dfrac{5}{3} > 0\\
\Rightarrow {x^2} + 2.x.\dfrac{1}{6} + \dfrac{1}{{36}} + \dfrac{{59}}{{36}} > 0\\
\Rightarrow {\left( {x + \dfrac{1}{6}} \right)^2} + \dfrac{{59}}{{36}} > 0\left( {tm} \right)\\
Vậy\,x \in R\\
c) - 9 - 2x - 7{x^2} < 0\\
\Rightarrow 7{x^2} + 2x + 9 > 0\\
\Rightarrow {x^2} + 2.\dfrac{1}{7}.x + \dfrac{9}{7} > 0\\
\Rightarrow {\left( {x + \dfrac{1}{7}} \right)^2} + \dfrac{{62}}{{49}} > 0\left( {tm} \right)\\
Vậy\,x \in R\\
2)\left| {3{x^2} - 4x - 7} \right| < 2\\
\Rightarrow - 2 < 3{x^2} - 4x - 7 < 2\\
\Rightarrow \left\{ \begin{array}{l}
3{x^2} - 4x - 5 > 0\\
3{x^2} - 4x - 9 < 0
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
\left[ \begin{array}{l}
x > \dfrac{{2 + \sqrt {19} }}{3}\\
x < \dfrac{{2 - \sqrt {19} }}{3}
\end{array} \right.\\
\left[ {\dfrac{{2 - \sqrt {31} }}{3} < x < \dfrac{{2 + \sqrt {31} }}{3}} \right.
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
\dfrac{{2 - \sqrt {31} }}{3} < x < \dfrac{{2 - \sqrt {19} }}{3}\\
\dfrac{{2 + \sqrt {19} }}{3} < x < \dfrac{{2 + \sqrt {31} }}{3}
\end{array} \right.
\end{array}$
