a, \(\frac{n+3}{n}=\frac{n}{n}+\frac{3}{n}=1+\frac{3}{n}\)
=> n thuộc Ư(3) = {1;-1;3;-3}
Vậy n thuộc {1;-1;3;-3}
b, \(\frac{2n+1}{n-1}=\frac{2n-2+3}{n-1}=\frac{2\left(n-1\right)+3}{n-1}=\frac{2\left(n-1\right)}{n-1}+\frac{3}{n-1}=2+\frac{3}{n-1}\)
=> n - 1 thuộc Ư(3) = {1;-1;3;-3}
=> n thuộc {2;0;4;-2}
Vậy n thuộc {2;0;4;-2}
a) n+3 \(⋮\) n
\(\Leftrightarrow\) 3\(⋮\)n ( vì n/n)
\(\Leftrightarrow\) n = \(\left\{-1;-3;1;3\right\}\)
b) 2n + 1 \(⋮\)n-1
\(\Leftrightarrow\)2n - 2 + 3\(⋮\)n-1
\(\Leftrightarrow\)2(n-1)+3\(⋮\)n-1
\(\Leftrightarrow\)3\(⋮\)n-1
\(\Leftrightarrow\)n-1 = \(\left\{-1;-3;1;3\right\}\)
\(\Leftrightarrow\)n= \(\left\{0;-2;2;4\right\}\)