tim n thuoc z a, n.n-7 chia het cho n+3 b,n+3 chia het cho n.n-7

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1. `a)` – Ta có : `n^2-7 vdots n+3`

   mà `3(n+3) vdots n+3`

   `=> n^2-7+3(n+3) vdots n+3`

   `=> n^2-7+3n+9 vdots n+3`

   `=> n(n+3)+(-7+9) vdots n+3`

   `=> n(n+3)-2 vdots n+3` 

   mà `n(n+3) vdots n+3`

   `=> -2 vdots n+3`

   `=> n+3 in Ư(-2)={+-1;+-2}`

   – Ta có bảng sau :

   $\begin{array}{|c|c|c|c|} \hline n+3&-2&-1&1&2 \\\hline n&-5&-4&-2&-1 \\\hline \end{array}$

   – Vậy `n in {-5;-4;-2;-1}`

   – Vậy `n in {-5;-4;-2;-1}`

    

   `b)` – Ta có : `n+3 vdots n^2-7`

   `=> (n+3)(n-3) vdots n^2-7`

   `=> n(n+3)-3(n+3) vdots n^2-7`

   `=> n^2+3n-(3n+9) vdots n^2-7`

   `=> n^2+(3n-3n)-9 vdots n^2-7`

   `=> n^2-9 vdots n^2-7`

   `=> n^2-7-2 vdots n^2-7`

   mà `n^2-7 vdots n^2-7`

   `=> -2 vdots n^2-7`

   `=> n.n-7 in Ư(-2)={+-1;+-2}`

   – Ta có bảng sau :

   $\begin{array}{|c|c|c|c|} \hline n^2-7&-2&-1&1&2 \\\hline n^2&5&6&8&9 \\\hline n&\text{loại}&\text{loại}&\text{loại}&\pm3 \\\hline \end{array}$ 

   – Vậy `n = +-3`

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