Bài 1:
$A=(n-1)(2n-3)-2n(n-3)-4n$
$=2n^2-5n+3-(2n^2-6n)-4n$
$=-3n+3=3(1-n)$ chia hết cho $3$ với mọi số nguyên $n$
Ta có đpcm.
Bài 2:
$B=(n+2)(2n-3)+n(2n-3)+n(n+10)$
$=(2n-3)(n+2+n)+n(n+10)$
$=(2n-3)(2n+2)+n(n+10)=4n^2-2n-6+n^2+10n$
$=5n^2+8n-6=5n(n+3)-7(n+3)+15$
$=(n+3)(5n-7)+15$
Để $B\vdots n+3$ thì $(n+3)(5n-7)+15\vdots n+3$
$\Leftrightarrow 15\vdots n+3$
$\Leftrightarrow n+3\in\left\{\pm 1;\pm 3;\pm 5;\pm 15\right\}$
$\Rightarrow n\in\left\{-2;-4;0;-6;-8; 2;12;-18\right\}$