Question

1.tìm n∈N biết :

b)n\(^2\)+3⋮n+1

Answer 1

\(\frac{n^2+3}{n+1}=\frac{n\left(n+1\right)-n+3}{n+1}=\frac{n\left(n+1\right)}{n+1}-\frac{n+3}{n+1}=n-\frac{n+3}{n+1}\in Z\)

Suy ra \(n+3⋮n+1\)\(\Leftrightarrow\frac{n+3}{n+1}=\frac{n+1+2}{n+1}=\frac{n+1}{n+1}+\frac{2}{n+1}=1+\frac{2}{n+1}\in Z\)

Suy ra \(2⋮n+1\)\(\Rightarrow n+1\inƯ\left(2\right)=\left\{1;-1;2;-2\right\}\)

\(\Rightarrow n\in\left\{0;1\right\}\left(n\in N\right)\)