a) Ta có: \(n^2+4⋮n+2\) (I)
Mà \(n+2⋮n+2\)
\(\Rightarrow n\left(n+2\right)⋮n+2\)
\(\Rightarrow n^2+2n⋮n+2\) (II)
Từ (I) và (II) \(\Rightarrow\left(n^2+2n\right)-\left(n^2+4\right)⋮n+2\)
\(\Rightarrow2n-4⋮n+2\)
\(\Rightarrow\left(2n+4\right)-8⋮n+2\)
\(\Rightarrow2\left(n+2\right)-8⋮n+2\)
\(\Rightarrow-8⋮n+2\)
\(\Rightarrow n+2\in\left\{1;2;4;8\right\}\) ( vì \(n\in N\) )
\(\Rightarrow\left\{{}\begin{matrix}n+2=1\Rightarrow n=-1\left(loai\right)\\n+2=2\Rightarrow n=0\\n+2=4\Rightarrow n=2\\n+2=8\Rightarrow n=6\end{matrix}\right.\)
Vậy n=0 hoặc n=2 hoặc n=6
b) Ta có: \(13n⋮n-1\)
\(\Rightarrow\left(13n-13\right)+13⋮n-1\)
\(\Rightarrow13\left(n-1\right)+13⋮n-1\)
\(\Rightarrow13⋮n-1\)
\(\Rightarrow n-1\in\left\{1;13\right\}\) ( vì \(n\in N\) )
\(\Rightarrow\left\{{}\begin{matrix}n-1=1\Rightarrow n=2\\n-1=13\Rightarrow n=14\end{matrix}\right.\)
Vậy n=2 hoặc n=14
Nguyễn Nam Giải dài dòng quá
a)\(n^2+4⋮n+2\)
\(\Rightarrow n^2-4+8⋮n+2\)
\(\Rightarrow\left(n+2\right)\left(n-2\right)+8⋮n+2\)
\(\Rightarrow8⋮n+2\)
b)
\(13n⋮n-1\)
\(\Rightarrow13n-13+13⋮n-1\)
\(\Rightarrow13⋮n-1\)
a) n2+4⋮ n+2
n2-4+8⋮ n+2
(n+2)(n-2)+8⋮ n+2
=>8⋮n+2
