Question

Tìm n > 0 để : n² + 2n + 5 chia hết cho n + 4

Answer 1

Ta có: \(\dfrac{n^2+2n+5}{n+4}=\dfrac{n\left(n+4\right)-\left(2n-5\right)}{n+4}\)

\(=\dfrac{n\left(n+4\right)}{n+4}-\dfrac{2n-5}{n+4}=n-\dfrac{2\left(n+4\right)-13}{n+4}\)

\(=n-\dfrac{2\left(n+4\right)}{n+4}-\dfrac{13}{n+4}=n-2-\dfrac{13}{n+4}\)

Để \(n^2+2n+5⋮n+4\)

Thì \(13⋮n+4\)\(\Rightarrow n+4\inƯ\left(13\right)=\left\{1;-1;13;-13\right\}\)

\(\Rightarrow n=9\left(n>0\right)\)

Answer 2

Ace Legona Hoàng Ngọc Anh Hồng Phúc Nguyễn Adonis Baldric Khánh Linh Nguyễn Thị Huyền Trang Nguyễn Huy Tú kudo shinichi Trần Thiên Kim Toshiro Kiyoshi ..... giúp mk nha