Question

Tìm số nguyên n sao cho n^2 + 3 chia hết cho n + 1

Answer 1

Giải:

Ta có: \(\dfrac{n^2+3}{n+1}\) 

\(=\dfrac{n^2+n-n-1+4}{n+1}\)

\(=\dfrac{n\left(n+1\right)-\left(n+1\right)+4}{n+1}\)

\(=\dfrac{n\left(n+1\right)}{n+1}-\dfrac{n+1}{n+1}+\dfrac{4}{n+1}=n-1+\dfrac{4}{n+1}\)

Để \(n^2+3⋮x+1\) thì \(4⋮n+1\) 

\(\Rightarrow n+1\inƯ\left(4\right)=\left\{\pm1;\pm2\pm4\right\}\) (đk: \(n\ne-1\))

\(\Rightarrow n\in\left\{-5;-3;-2;0;1;3\right\}\) (t/m)

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