n+1\(⋮\)7 

\(\Rightarrow\)5n+1+14\(⋮7\)

\(\Rightarrow5n+15⋮7\)

\(\Rightarrow5(n+3)⋮7\)

\(\Rightarrow n+3⋮7\left(vi(5:7)=1\right)\)

\(\Rightarrow n+3\in B_{\left(7\right)}\)

\(\Rightarrow n+3=7k\left(k\inℕ^∗\right)\)

\(\Rightarrow n=7k-3\)

vậy n có dạng 7k-3

a, Ta có : 2n + 19 chia hết cho 7

\(\Rightarrow\) \(2n+19\inƯ\left(7\right)\)

Mà  \(Ư\left(7\right)=\left\{1;-1;7;-7\right\}\)

\(\Rightarrow\) \(2n+19\in\left\{1;-1;7;-7\right\}\)

\(\Rightarrow\) \(2n\in\left\{20;18;26;12\right\}\)

\(\Rightarrow\) \(n\in\left\{10;9;13;6\right\}\)

1/

$10n+4\vdots 2n+7$

$\Rightarrow 5(2n+7)-31\vdots 2n+7$

$\Rightarrow 31\vdots 2n+7$

$\Rightarrow 2n+7\in Ư(31)$

$\Rightarrow 2n+7\in \left\{1; -1; 31; -31\right\}$

$\Rightarrow n\in \left\{-3; -4; 12; -19\right\}$

2/

$5n-4\vdots 3n+1$

$\Rightarrow 3(5n-4)\vdots 3n+1$

$\Rightarroq 15n-12\vdots 3n+1$

$\Rightarrow 5(3n+1)-17\vdots 3n+1$

$\Rightarrow 17\vdots 3n+1$

$\Rightarrow 3n+1\in Ư(17)$

$\Rightarrow 3n+1\in \left\{1; -1; 17; -17\right\}$

$\Rightarrow n\in \left\{0; \frac{-2}{3}; \frac{16}{3}; -6\right\}$

Do $n$ nguyên nên $n\in\left\{0; -6\right\}$

Đề có đúng ko vậy bn