Đáp án:
$\begin{array}{l}
1)\\
{x^3} - \left( {2 + 3x} \right)\left( {3x - 2} \right) = \left( {4 - 2x + {x^2}} \right)\left( {x + 2} \right)\\
- \left( {3x + 5} \right)\\
\Rightarrow {x^3} - \left( {9{x^2} - 4} \right) = {x^3} + 8 - 3x - 5\\
\Rightarrow {x^3} - 9{x^2} + 4 = {x^3} - 3x + 3\\
\Rightarrow 9{x^2} - 3x - 1 = 0\\
\Rightarrow {\left( {3x} \right)^2} - 2.3x.\dfrac{1}{2} + \dfrac{1}{4} = \dfrac{5}{4}\\
\Rightarrow {\left( {3x - \dfrac{1}{2}} \right)^2} = \dfrac{5}{4}\\
\Rightarrow \left[ \begin{array}{l}
3x - \dfrac{1}{2} = \dfrac{{\sqrt 5 }}{2}\\
3x - \dfrac{1}{2} = - \dfrac{{\sqrt 5 }}{2}
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
x = \dfrac{{3\sqrt 5 + 3}}{2}\\
x = \dfrac{{ - 3\sqrt 5 + 3}}{2}
\end{array} \right.\\
2)\\
2 + 4{x^2} - 3x\\
= 4{x^2} - 3x + 2\\
= {\left( {2x} \right)^2} - 2.2x.\dfrac{3}{4} + \dfrac{9}{{16}} + \dfrac{{23}}{{16}}\\
= {\left( {2x - \dfrac{3}{4}} \right)^2} + \dfrac{{23}}{{16}} \ge \dfrac{{23}}{{16}} > 0\\
Vay\,2 + 4{x^2} - 3x > 0
\end{array}$
