1/

`8n+3\vdots6n-1`

`⇔48n+18\vdots6n-1`

`⇔48n+18-8(6n-1)vdots6n-1`

`⇔26\vdots6n-1`

`⇔6n-1∈Ư(26)∈{±1;±2;±13;±26}`

`⇔6n∈{2;0;14;-12;3;-1;27;-25}`

mà `n∈Z`

`⇒n{0;-2}`

2/

`n(2n+2)(5n+10)\vdots 60`

`⇔n.(n+1).2.(n+2).5\vdots2;3;5`

Ta có:

`n.(n+1).2.(n+2).5\vdots2;5`

mà `n(n+1)(n+2)` là ba số tự nhiên liên tiếp nên có `1` số chia hết cho `3`

`⇒n.(n+1).2.(n+2).5\vdots2;3;5`

`⇒n(2n+2)(5n+10)\vdots 60`

 

Bài 1:

Ta có: $\left \{ {{8n+3 \vdots 6n-1} \atop {6n-1\vdots 6n-1}} \right.$ 

 `<=>` $\left \{ {{6(8n+3)\vdots 6n-1 } \atop {8(6n-1)\vdots 6n-1}} \right.$ 

`=>6(8n+3)-8(6n-1)\vdots 6n-1`

`=>48n+18-48n+8 \vdots 6n-1`

`=>26 \vdots 6n-1`

`=>6n-1∈Ư(26)={+-1;+-2;+-13;+-26}`

`=>6n={0;-1;-12;-25;2;3;14;27}`

`=>n={0;-1/6;-2;-25/6;1/3;1/2;7/3;9/2}`

Mà: `n∈Z` nên:

`=>n={0;-2}`

Bài 2:

Ta có: `n(2n+2)(5n+10)\vdots60 `

Hay: `n(2n+2)(5n+10) \vdots 2*3*5*2`

`n(2n+2)(5n+10)\vdots60 `

`<=>n(2n+2)(5n+10) \vdots 2*3*5`

`n(2n+2)(5n+10)`

`=2n(n+1)5(n+2)`

Ta có: `5 \vdots 5=> 5(n+2) \vdots 5(***)`

Và: `n(n+1)(n+2)` là `3` số tự nhiên liên tiếp nên:

`=>n(n+1)(n+2) \vdots 2;3(******)` 

Từ: `(***)+(******)=>n(2n+2)(5n+10)\vdots60`