Giải thích các bước giải:
a.Ta có:
$3n+13\quad\vdots\quad n+1$
$\to (3n+3)+10\quad\vdots\quad n+1$
$\to 3(n+1)+10\quad\vdots\quad n+1$
$\to 10\quad\vdots\quad n+1$
$\to n+1\in\{1,2,5,10\}$ vì $n\in N$
$\to n\in\{0,1,4,9\}$
b.Ta có:
$2n+7\quad\vdots\quad3n+1$
$\to 3(2n+7)\quad\vdots\quad3n+1$
$\to 6n+21\quad\vdots\quad3n+1$
$\to 6n+2+19\quad\vdots\quad3n+1$
$\to 2(3n+1)+19\quad\vdots\quad3n+1$
$\to 19\quad\vdots\quad3n+1$
$\to 3n+1$ là ước của $19$
Mà $n\ge 0\to 3n+1\ge 1$
$\to 3n+1\in\{1,19\}$
$\to 3n\in\{0,18\}$
$\to n\in\{0,6\}$
c.Ta có:
$n^2+3n-13\quad\vdots\quad n+3$
$\to (n^2+3n)-13\quad\vdots\quad n+3$
$\to n(n+3)-13\quad\vdots\quad n+3$
$\to 13\quad\vdots\quad n+3$
$\to n+3=13$ vì $n+3\ge 0+3=3$
$\to n=10$
d.Ta có:
$n^2+3\quad\vdots\quad n-1$
$\to (n^2-1)+4\quad\vdots\quad n-1$
$\to (n-1)(n+1)+4\quad\vdots\quad n-1$
$\to 4\quad\vdots\quad n-1$
$\to n-1\in\{-1,1,2,4\}$ vì $n\in N\to n\ge 0\to n-1\ge -1$
$\to n\in\{0,2,3,5\}$
