a)
\(n+4⋮n+1\Leftrightarrow\left(n+1\right)+3⋮n+1\)
\(3⋮n+1\)(vì n+1 chia hết cho n+1)
\(\Rightarrow n+1\inƯ\left(3\right)=\left\{1;3\right\}\)
\(n+1=1\Rightarrow n=0\)
\(n+1=3\Rightarrow n=2\)
Vậy \(n\in\left\{0;2\right\}\)
b)
\(2n+3⋮n+1\Leftrightarrow2\left(n+1\right)+1⋮n+1\)
\(\Rightarrow1⋮n+1\)(vì 2(n+1) chia hết cho n+1)
\(\Rightarrow n+1\inƯ\left(1\right)=\left\{1\right\}\)
\(\Rightarrow n+1=1\Rightarrow n=0\)
Vậy \(n=0\)
ta có 4n+ 7 chia hết cho 2n +1 (1)
2n+ 1 chia hết cho 2n+1
=> 2(2n+1) chia hết cho 2n+1
=> 4n+2 chia hết cho 2n+1 (2)
từ (1) và (2)