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`
