`=>`
$a)$ `4n-5\vdots 13`
`=> 4n-5+13\vdots 13`
`=> 4n+8\vdots 13`
`=> 4(n+2) \vdots 13`
`=> n+2\vdots 13` (vì `(4; 13)=1)`
`=> n+2=13k (k\in NN^**)`
`=> n=13k-2`
Vậy `n` có dạng là `13k-2` khi `4n-5 \vdots 13`.
$b)$ `5n+1 \vdots 7`
`=> 5n+1+14 \vdots 7`
`=> 5n+15 \vdots 7`
`=> 5(n+3) \vdots 7`
`=> n+3 \vdots 7`
`=> n+3 \in B(7)` (vì `(5; 7)=1)`
`=> n+3=7k (k\in NN^**)`
`=> n=7k-3`
Vậy `n` có dạng là `7k-3` khi `5n+1\vdots 7`.
$c)$ `25n+3\vdots 53`
`=> 25n+3-53\vdots 53`
`=> 25n-50\vdots 53`
`=> 25n(n-2)\vdots 53`
`=> n-2\vdots 53` (vì `(25; 53=1)`
`=> n-2=53k (k\in NN^**)`
`=> n=53k+2`
Vậy `n` có dạng `53k+2` khi `25n+3 \vdots 53`.
Giải thích:
$a)$ `=> 4n-5+13\vdots 13`
`=> 4n+8\vdots 13`
`=> 4(n+2) \vdots 13`
`-> 4n-5+13\vdots 13`
`-> 4n+(-5)+13\vdots 13`
`-> 4n+8\vdots 13`
`-> 4n+4.2\vdots 13`
`-> 4(n+2)\vdots 13`
$_{*}$ $\text{b, c tương tự.}$
