tìm số nguyên n biết (n^2 + 3 ) chia hết cho (n+1)

Question

tìm số nguyên n biết
(n^2 + 3 ) chia hết cho (n+1)

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Isabelle 2 tháng 2021-10-08T14:17:37+00:00 2 Answers 5 views 0

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    0
    2021-10-08T14:18:53+00:00

    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}`

    0
    2021-10-08T14:19:20+00:00

    $\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}$

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