Giải thích các bước giải:
Bài 2:
1) `(2n + 1) \vdots (6n - 1)`
`⇒ 3(2n + 1) - (6n - 1) \vdors (6n-1)`
`⇒ 4 \vdots (6n - 1)`
`⇒ 6n - 1 ∈ Ư(4) = {±1;±2±4}`
mà `6n - 1` lẻ `=> 6n - 1 = ±1`
Ta có bảng sau:
$\begin{array}{|c|c|}\hline 6n-1&-1&1\\ \hline n &0&/ \\ \hline \end{array}$
Vậy ` n = 0` thì `(2n + 1) \vdots (6n-1)`
__________
2) `3n \vdots (5 - 2n)`
`=> 6n \vdots (5-2n)`
`=> 6n + 3(5 - 2n) \vdots (5-2n)`
`=> 15 \vdots (5-2n) ⇒ 5 - 2n ∈ Ư(15) = {±1;±3;±5;±15}`
Ta có bảng sau:
$\begin{array}{|c|c|}\hline 5-2n&-15&-5&-3&-1&1&3&5&15 \\\hline n &10&5&4&3&2&1&0&-5\\\hline\end{array}$
Vậy `n ∈ {10; ±5; 4;3;2;1;0}` thì `3n \vdots (5-2n)`
_________
3) `(n^2 + 1) \vdots (n-3)`
`=> n^2 + 1 - n(n-3) \vdots (n-3)`
`=> 1 + 3n \vdots (n-3)`
`=> 1 + 3n - 3(n-3) \vdots (n-3)`
`=> 10 \vdots (n-3)⇒ n - 3 ∈ Ư(10) = {±1;±2;±5;±10}`
Ta có bảng sau:
$\begin{array}{|c|c|}\hline n-3&-10&-5&-2&-1&1&2&5&10\\\hline n &-7&-2&1&2&4&5&8&13\\\hline \end{array}$
Vậy `n ∈ {-7;-2;1;2;4;5;8;13}` thì `(n^2 + 1) \vdots (n-3)`
__________
4) `(n^2 + 2n + 7) \vdots (n+2)`
`=> n^2+2n+7-n(n+2) \vdots (n+2)`
`=> 7 \vdots (n+2)=> n + 2 ∈Ư(7)={±1;±7}`
Ta có bảng sau:
$\begin{array}{|c|c|} \hline n+2&-7&-1&1&7\\\hline n &-9&-3&-1&5\\\hline \end{array}$
Vậy `n ∈ {-9; -3;-1;5}` thì `(n^2 + 2n + 7) \vdots (n+2)`
__________
5) `(8n+193)/(4n+3)=(2(4n+3)+187)/(4n+3)=(2(4n+3))/(4n+3)+187/(4n+3)=2 + 187/(4n+3)`
`2` nguyên, để `(8n + 193)/(4n+3) ∈ Z <=> 187/(4n + 3) ∈ Z`
`⇒ 4n+3 ∈ Ư(187)={±1;±11;±17;±187}`
Ta có bảng sau:
$\begin{array}{|c|c|} \hline4n+3&-187&-17&-11&-1&1&11&17&187\\\hline n&/&-5&/&-1&/&2&/&46\\\hline\end{array}$
Vậy `n ∈ {-5;-1;2;46}` thì `(8n+193)/(4n+3) ∈ Z`
