Giải thích các bước giải:

a.Ta có :
$4n+3\quad\vdots\quad n-2$ 

$\to 4n-8+11\quad\vdots\quad n-2$ 

$\to 4(n-2)+11\quad\vdots\quad n-2$ 

$\to 11\quad\vdots\quad n-2$ 
$\to n-2\in\{11,1,-1,-11\}$

$\to n\in\{13,3,1,-9\}$

b.Ta có :
$n+4\quad\vdots\quad n+1$ 

$\to n+1+3\quad\vdots\quad n+1$ 

$\to 3\quad\vdots\quad n+1$ 

$\to n+1\in\{1,3,-1,-3\}$

$\to n\in\{0,2,-2,-4\}$

c.Ta có :
$5(6x+11y)+(x+7y)=31x+62y=31(x+2y)\quad\vdots\quad 31$

Mà $6x+11y\quad\vdots\quad 31\to x+7y\quad\vdots\quad 31$

a. \[4n+3\quad\vdots\quad n-2\]

\[\to 4n-8+11\quad\vdots\quad n-2\]

\[\to 4(n-2)+11\quad\vdots\quad n-2\]

Vì $n\in \mathbb{Z} \to 4(n-2)\quad\vdots\quad n-2$

\[\to 11\quad\vdots\quad n-2\]

\[\to n-2\in U_{\{11\}}=\{±1;±11\}\]

\[\to n\in\{3;1;13;-9\}\]

b. \[n+4\quad\vdots\quad n+1\]

\[\to (n+1)+3\quad\vdots\quad n+1\]

Vì $n\in \mathbb{Z} \to n(n+1)\quad\vdots\quad n+1$

\[\to 3\quad\vdots\quad n+1\]

\[\to n+1∈U_{\{3\}}=\{±1;±3\}\]

\[\to n∈\{0;-2;2;-4\}\]

c. 

\[6x+11y\quad\vdots\quad 31\] \[\to 5(6x+11y)\quad\vdots\quad 31\] \[\to 30x+55y\quad\vdots\quad 31\] mà \[31(x+2y)\quad\vdots\quad 31\] \[→31x+62y\quad\vdots\quad 31\] \[→(31x+62y)-(30x+55y)\quad\vdots\quad 31\] \[\to x+7y\quad\vdots\quad 31\]