tìm n sao cho $2n^{2}$ +n-7 chia hết cho n-2 02/12/2021 Bởi Katherine tìm n sao cho $2n^{2}$ +n-7 chia hết cho n-2
Có `2n^2+n-7 vdots n-2` ( `n \ne 2`) `<=>(2n^2+n-7)/(n-2) in Z` Mà `(2n^2+n-7)/(n-2)=2n+5+3/(n-2)` Có `2n+5 in Z ` khi `n in Z` Có `2n^2+n-7 vdots n-2` `<=>(2n^2+n-7)/(n-2) in Z` `<=>2n+5+3/(n-2) in Z` `<=>3/(n-2) in Z` `<=>n-2 in Ư(3)={1;-1;-3;3}` `<=> n in {3;1;-1;5}` Bình luận
IDK
Có
`2n^2+n-7 vdots n-2` ( `n \ne 2`)
`<=>(2n^2+n-7)/(n-2) in Z`
Mà `(2n^2+n-7)/(n-2)=2n+5+3/(n-2)`
Có `2n+5 in Z ` khi `n in Z`
Có
`2n^2+n-7 vdots n-2`
`<=>(2n^2+n-7)/(n-2) in Z`
`<=>2n+5+3/(n-2) in Z`
`<=>3/(n-2) in Z`
`<=>n-2 in Ư(3)={1;-1;-3;3}`
`<=> n in {3;1;-1;5}`