Giải thích các bước giải:
$\begin{array}{l}
1){x^2} + 3x + m = \left( {2x - 1} \right)\left( {\frac{1}{2}x + a} \right) = {x^2} + \left( {2a - \frac{1}{2}} \right)x - a\\
\Rightarrow \left\{ \begin{array}{l}
2a - \frac{1}{2} = 3\\
- a = m
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
a = \frac{7}{4}\\
m = \frac{{ - 7}}{4}
\end{array} \right.
\end{array}$
Vậy $m = \frac{{ - 7}}{4}$
$2){x^3} - 5x - 6 = \left( {{x^2} - x + 1} \right)\left( {x + 1} \right) - 5x - 7$
Vậy đa thức dư là: $-5x-7$
$\begin{array}{l}
3)2{x^2} + 3x + 3 = \left( {2x - 1} \right)\left( {x + 2} \right) + 5\\
\left( {2{x^2} + 3x + 3} \right) \vdots \left( {2x - 1} \right) \Leftrightarrow 5 \vdots \left( {2x - 1} \right)\\
\Leftrightarrow \left( {2x - 1} \right) \in U\left( 5 \right)\left( {x \in Z} \right)\\
\Leftrightarrow \left( {2x - 1} \right) \in \left\{ { - 5; - 1;1;5} \right\}\\
\Leftrightarrow x \in \left\{ { - 2;0;1;3} \right\}
\end{array}$
Vậy $x \in \left\{ { - 2;0;1;3} \right\}$
