Toán tìm số nguyên n biết (n^2 + 3 ) chia hết cho (n+1) 08/10/2021 By Isabelle tìm số nguyên n biết (n^2 + 3 ) chia hết cho (n+1)
Đáp án: Ta có : `(n^2+3)/(n+1) = (n.n+3)/(n+1) = (n.(n+1)+2)/(n+1) ` mà `n(n+1) ⋮ n+1` `-> 2 ⋮ n+1` `-> n+1 ∈ {±1 ; ±2}` `-> n ∈ {0,-2;1,-3}` Vậy `n ∈ {0,-2;1,-3}` Trả lời
$\begin{array}{l}n^2+3\ \vdots\ n+1\\\Leftrightarrow n^2+3-n(n+1)\ \vdots\ n+1\\\Leftrightarrow n^2+3-n^2-n\ \vdots\ n+1\\\Leftrightarrow 3-n\ \vdots\ n+1\\\Leftrightarrow 3-n+(n+1)\ \vdots\ n+1\\\Leftrightarrow 3-n+n+1\ \vdots\ n+1\\\Leftrightarrow 4\ \vdots\ n+1\\\Leftrightarrow n+1\in Ư(4)=\{\pm1;\pm2;\pm4\}\\\text{- Ta có bảng sau :}\\\begin{array}{|c|c|}\hline n+1&-4&-2&-1&1&2&4\\\hline n&-5&-3&-2&0&1&3\\\hline\end{array}\\\text{- Vậy $n\in\{-5;-3;-2;0;1;3\}$} \end{array}$ Trả lời
Đáp án:
Ta có :
`(n^2+3)/(n+1) = (n.n+3)/(n+1) = (n.(n+1)+2)/(n+1) `
mà `n(n+1) ⋮ n+1`
`-> 2 ⋮ n+1`
`-> n+1 ∈ {±1 ; ±2}`
`-> n ∈ {0,-2;1,-3}`
Vậy `n ∈ {0,-2;1,-3}`
$\begin{array}{l}n^2+3\ \vdots\ n+1\\\Leftrightarrow n^2+3-n(n+1)\ \vdots\ n+1\\\Leftrightarrow n^2+3-n^2-n\ \vdots\ n+1\\\Leftrightarrow 3-n\ \vdots\ n+1\\\Leftrightarrow 3-n+(n+1)\ \vdots\ n+1\\\Leftrightarrow 3-n+n+1\ \vdots\ n+1\\\Leftrightarrow 4\ \vdots\ n+1\\\Leftrightarrow n+1\in Ư(4)=\{\pm1;\pm2;\pm4\}\\\text{- Ta có bảng sau :}\\\begin{array}{|c|c|}\hline n+1&-4&-2&-1&1&2&4\\\hline n&-5&-3&-2&0&1&3\\\hline\end{array}\\\text{- Vậy $n\in\{-5;-3;-2;0;1;3\}$} \end{array}$