Đáp án:
$\begin{array}{l}
1)a)A = \left( {5x - 1} \right)\left( {x + 3} \right) - \left( {x - 2} \right)\left( {5x - 4} \right)\\
= 5{x^2} + 15x - x - 3 - \left( {5{x^2} - 4x - 10x + 8} \right)\\
= 5{x^2} + 14x - 3 - \left( {5{x^2} - 14x + 8} \right)\\
= 28x - 11\\
b)\\
B = \left( {3a - 2b} \right)\left( {9{a^2} + 6ab + 4{b^2}} \right)\\
= {\left( {3a} \right)^3} - {\left( {2b} \right)^3}\\
= 27{a^3} - 8{b^3}\\
2)\\
n\left( {2n - 3} \right) - 2n\left( {n + 2} \right)\\
= 2{n^2} - 3n - 2{n^2} - 4n\\
= - 7n \vdots 7\left( {khi:n \in Z} \right)\\
3)\\
{x^4} - 3x + 2\\
= \left( {x - 1} \right)\left( {{x^3} + b{x^2} + a.x - 2} \right)\\
= {x^4} + b{x^3} + a.{x^2} - 2x - {x^3} - b{x^2} - a.x + 2\\
= {x^4} + \left( {b - 1} \right).{x^3} + \left( {a - b} \right).{x^2} - \left( {a + 2} \right).x + 2\\
\Leftrightarrow \left\{ \begin{array}{l}
b - 1 = 0\\
a - b = 0\\
a + 2 = 3
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
b = 1\\
a = b\\
a = 1
\end{array} \right. \Leftrightarrow a = b = 1\\
Vậy\,a = b = 1\\
4)\\
n\left( {n + 5} \right) - \left( {n - 3} \right)\left( {n + 2} \right)\\
= {n^2} + 5n - \left( {{n^2} + 2n - 3n - 6} \right)\\
= {n^2} + 5n - {n^2} + n + 6\\
= 6n + 6\\
= 6.\left( {n + 1} \right) \vdots 6\left( {khi:n \in Z} \right)
\end{array}$
