$n^{2}$ +13n-13 chia hết cho n+3 Tìm n ∈ Z 07/10/2021 Bởi Kinsley $n^{2}$ +13n-13 chia hết cho n+3 Tìm n ∈ Z
Đáp án: `n∈\{-46;-4;-2;40\}` Giải thích các bước giải: `n^2+13n-13` `=n^2+3n+10n+30-43` `=n(n+3)+10(n+3)-43` Vì `n^2+13n-13\vdots (n+3)` `\Rightarrow n(n+3)+10(n+3)-43\vdots (n+3)` `\Rightarrow 43\vdots (n+3)(`Do`n(n+3)\vdots (n+3);10(n+3)\vdots (n+3))` `\Rightarrow (n+3)∈Ư(43)=\{-43;-1;1;43\}` `\Rightarrow n∈\{-46;-4;-2;40\}` Bình luận
Để `(n^2+13n-13)\ \vdots \ (n+3)` `=>(n^2+3n+10n+30-43)\ \vdots \ (n+3)` `=>n(n+3)+10.(n+3)-43\ \ \vdots \ (n+3)` `=>(n+3).(n+10)-43\ \ \vdots \ (n+3)` $\\$ Vì `(n+3)(n+10)\ \vdots \ (n+3)` `=>43\ \vdots \ (n+3)` `=>n+3\in Ư(43)={-43;-1;1;43}` `=>n\in {-46;-4;-2;40}` $\\$ Vậy `n\in {-46;-4;-2;40}` thỏa đề bài Bình luận
Đáp án:
`n∈\{-46;-4;-2;40\}`
Giải thích các bước giải:
`n^2+13n-13`
`=n^2+3n+10n+30-43`
`=n(n+3)+10(n+3)-43`
Vì `n^2+13n-13\vdots (n+3)`
`\Rightarrow n(n+3)+10(n+3)-43\vdots (n+3)`
`\Rightarrow 43\vdots (n+3)(`Do`n(n+3)\vdots (n+3);10(n+3)\vdots (n+3))`
`\Rightarrow (n+3)∈Ư(43)=\{-43;-1;1;43\}`
`\Rightarrow n∈\{-46;-4;-2;40\}`
Để `(n^2+13n-13)\ \vdots \ (n+3)`
`=>(n^2+3n+10n+30-43)\ \vdots \ (n+3)`
`=>n(n+3)+10.(n+3)-43\ \ \vdots \ (n+3)`
`=>(n+3).(n+10)-43\ \ \vdots \ (n+3)`
$\\$
Vì `(n+3)(n+10)\ \vdots \ (n+3)`
`=>43\ \vdots \ (n+3)`
`=>n+3\in Ư(43)={-43;-1;1;43}`
`=>n\in {-46;-4;-2;40}`
$\\$
Vậy `n\in {-46;-4;-2;40}` thỏa đề bài