Tìm số nguyên n để n^2+n+1 chia hết cho n+3 03/07/2021 Bởi Ruby Tìm số nguyên n để n^2+n+1 chia hết cho n+3
$\dfrac{n^2+n+1}{n+3}(n\ne 3)\\=\dfrac{n^2+3n-2n+1}{n+3}\\=\dfrac{n(n+3)-2n-6+7}{n+3}\\=n-\dfrac{2(n+3)+7}{n+3}\\=n-2+\dfrac{7}{n+3}$ Để $n^2+n+1\vdots n+3$ $→\begin{cases}n∈\Bbb Z\\7\vdots n+3\end{cases}$ $7\vdots n+3\\→n+3∈Ư(7)=\{±1;±7\}$ $→$ Ta có bảng: $\begin{array}{|c|c|c|}\hline n+3&1&-1&7&-7\\\hline n&-2&-4&4&-10\\\hline \quad&tm&tm&tm&tm\\\hline\end{array}$ $⇒n∈\{-2;-4;4;-10\}$ Vậy $n∈\{-2;-4;4;-10\}$ Bình luận
Đáp án: `n^2+n+1 vdots n+3` `<=>n^2+3n-2n+1 vdots n+3` `<=>n(n+3)-2n+1 vdots n+3` `<=>-2n+1 vdots n+3` `<=>-2(n+3)+7 vdots n+3` `<=>7 vdots n+3` `<=>n+3 in Ư(7)={+-1,+-7}` `<=>n in {-4,-2,4,-10}`. Vậy `n in {-4,-2,4,-10}` thì `n^2+n+1 vdots n+3` Bình luận
$\dfrac{n^2+n+1}{n+3}(n\ne 3)\\=\dfrac{n^2+3n-2n+1}{n+3}\\=\dfrac{n(n+3)-2n-6+7}{n+3}\\=n-\dfrac{2(n+3)+7}{n+3}\\=n-2+\dfrac{7}{n+3}$
Để $n^2+n+1\vdots n+3$
$→\begin{cases}n∈\Bbb Z\\7\vdots n+3\end{cases}$
$7\vdots n+3\\→n+3∈Ư(7)=\{±1;±7\}$
$→$ Ta có bảng:
$\begin{array}{|c|c|c|}\hline n+3&1&-1&7&-7\\\hline n&-2&-4&4&-10\\\hline \quad&tm&tm&tm&tm\\\hline\end{array}$
$⇒n∈\{-2;-4;4;-10\}$
Vậy $n∈\{-2;-4;4;-10\}$
Đáp án:
`n^2+n+1 vdots n+3`
`<=>n^2+3n-2n+1 vdots n+3`
`<=>n(n+3)-2n+1 vdots n+3`
`<=>-2n+1 vdots n+3`
`<=>-2(n+3)+7 vdots n+3`
`<=>7 vdots n+3`
`<=>n+3 in Ư(7)={+-1,+-7}`
`<=>n in {-4,-2,4,-10}`.
Vậy `n in {-4,-2,4,-10}` thì `n^2+n+1 vdots n+3`