Đáp án:
`a) 3n+17`$\vdots$ `n+3`
`<=> 3n+9+8`$\vdots$`n+3`
`<=> 3(n+3)+8`$\vdots$`n+3`
mà `3(n+3)`$\vdots$ `n+3`
`<=> 8`$\vdots$ `n+3`
`<=> n+3∈Ư(8)={1;2;4;8}`
Do `n∈N=>n+3>=3`
`=> n+3∈{4;8}`
`=>n∈{1;5}`
Vậy `n∈{1;5}`
`b) n+9`$\vdots$ `2n+3`
`<=> 2(n+9)`$\vdots$ `2n+3`
`<=> 2n+18`$\vdots$ `2n+3`
`<=> 2n+3+15`$\vdots$ `2n+3`
mà `2n+3`$\vdots$ `2n+3`
`=> 15`$\vdots$ `2n+3`
`=> 2n+3∈Ư(15)={1;3;5;15}`
Do `n∈N=> 2n+3>=3`
`=> 2n+3∈{3;5;15}`
`=> 2n∈{0;2;12}`
`=>n∈{0;1;6}`
Vậy `n∈{0;1;6}`
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