Giải thích các bước giải:
B1:
Ta có:
$\begin{array}{l}
P\left( x \right) = 2{x^4} - 8{x^3} + 5{x^2} + mx + n\\
= {x^2}\left( {2{x^2} - 1} \right) - 4x\left( {2{x^2} - 1} \right) + 3\left( {2{x^2} - 1} \right) + \left( {m - 4} \right)x + n + 3\\
= \left( {2{x^2} - 1} \right)\left( {{x^2} - 4x + 3} \right) + \left( {m - 4} \right)x + n + 3
\end{array}$
Như vậy:
$\begin{array}{l}
P\left( x \right) \vdots \left( {2{x^2} - 1} \right)\\
\Leftrightarrow \left( {m - 4} \right)x + n + 3 = 0\\
\Leftrightarrow \left\{ \begin{array}{l}
m - 4 = 0\\
n + 3 = 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
m = 4\\
n = - 3
\end{array} \right.
\end{array}$
Vậy $m = 4;n = - 3$
B2:
$\begin{array}{l}
A = 10{x^4} - 13{x^3} - 9{x^2} + x + 18\\
= 5{x^3}\left( {2x - 3} \right) + {x^2}\left( {2x - 3} \right) - 3x\left( {2x - 3} \right) + 5\left( {2x - 3} \right) + 33\\
= \left( {2x - 3} \right)\left( {5{x^3} + {x^2} - 3x + 5} \right) + 33
\end{array}$
Như vậy:
$\begin{array}{l}
A \vdots B\\
\Leftrightarrow \left( {\left( {2x - 3} \right)\left( {5{x^3} + {x^2} - 3x + 5} \right) + 33} \right) \vdots \left( {2x - 3} \right)\\
\Leftrightarrow 33 \vdots \left( {2x - 3} \right)\\
\Leftrightarrow 2x - 3 \in U\left( {33} \right)\left( {Do:x \in Z} \right)\\
\Leftrightarrow 2x - 3 \in \left\{ { - 33; - 11; - 3; - 1;1;3;11;33} \right\}\\
\Leftrightarrow x \in \left\{ { - 15; - 4;0;1;3;7;18} \right\}
\end{array}$
Vậy $x \in \left\{ { - 15; - 4;0;1;3;7;18} \right\}$
