a) n + 3 ⋮n - 1
=>(n+3)-(n-1)⋮(n-1)
=> (n+3-n+1)⋮(n-1)
=> 4⋮(n-1)
=> n-1∈Ư(4)={1;2;4}
=> x∈{2;3;5}
vậy x∈{2;3;5}
2)
a) Ta có:
\(n+3⋮n-1\)
\(\Rightarrow\left(n-1\right)+4⋮n-1\)
\(\Rightarrow4⋮n-1\)
\(\Rightarrow n-1\in U\left(4\right)=\left\{1;2;4\right\}\) ( Vì \(n\in N\) )
\(\Rightarrow\left\{{}\begin{matrix}n-1=1\Rightarrow n=2\\n-1=2\Rightarrow n=3\\n-1=4\Rightarrow n=5\end{matrix}\right.\)
Vậy \(n\in\left\{2;3;5\right\}\)
b) Ta có:
\(4n+3⋮2n+1\)
\(\Rightarrow\left(4n+2\right)+1⋮2n+1\)
\(\Rightarrow2\left(2n+1\right)+1⋮2n+1\)
\(\Rightarrow1⋮2n+1\)
\(\Rightarrow2n+1\in U\left(1\right)=\left\{1\right\}\) ( Vì \(n\in N\) )
\(\Rightarrow2n+1=1\)
\(\Rightarrow n=0\)
Vậy \(n=0\)
