Đáp án:

 

Giải thích các bước giải:

`bb(a,` Ta có: `(2n+6)vdots(2n-1)`

`(2n+6)-(2n-1)vdots(2n-1)`

`2n+6-2n+1vdots(2n-1)`

`7vdots(2n-1)`

Suy ra `(2n-1)inƯ(7)={+-1;+-7}.`

Ta có bảng sau:

\begin{array}{|c|c|c|}\hline \text{2n - 1}&\text{-7}&\text{-1}&\text{1}&\text{7}\\\hline \text{n}&\text{-3}&\text{0}&\text{1}&\text{4}\\\hline\end{array}

Mà `n inNN`, vậy để `(2n+6)vdots(2n-1)` thì `bb(n in{0;1;4}.)`

`bb(b,)` Ta có: `(3n+7)vdots(n-2)`

`(3n+7)-3(n-2)vdots(n-2)`

`3n+7-3n+6vdots(n-2)`

`13vdots(n-2)`

Suy ra `(n-2)inƯ(13)={+-1;+-13}.`

\begin{array}{|c|c|c|}\hline \text{n - 2}&\text{-13}&\text{-1}&\text{1}&\text{13}\\\hline \text{n}&\text{-11}&\text{1}&\text{3}&\text{15}\\\hline\end{array}

Mà `n inNN`, vậy để `(3n+7)vdots(n-2)` thì `bb(n in{1;3;15}.`