a,Ta có:
2n+1\(⋮\)n+1
<=>2n+1-(n+1)\(⋮\)n+1
<=>2n+1-2(n+1)\(⋮\)n+1
<=>2n+1-2n-2\(⋮\)n+1
<=>-1\(⋮\)n+1
=>n+1\(\in\){-1;1}
=>n\(_{\in}\){-2;0}
Vậy n\(\in\){-2;0}
b,Ta có:
n-3\(⋮\)n+1
<=>n-3-(n+1)\(⋮\)n+1
<=>n-3-n-1\(⋮\)n+1
<=>-4\(⋮\)n+1
=>n+1\(\in\){-1;1;-2;2;-4;4}
=>n\(\in\){-2;0;-3;-5;3}
Vậy n\(\in\){-2;0;-3;-5;3}
\(a.\)\(2n+1⋮n+1\Leftrightarrow2n+2-1⋮n+1\Leftrightarrow2\left(n+1\right)-1⋮n+1\)
\(\Leftrightarrow1⋮n+1\)(vì \(2\left(n+1\right)⋮n+1\))
\(\Leftrightarrow n+1\inƯ\left(1\right)\Leftrightarrow n+1\in\left\{\pm1\right\}\Leftrightarrow n\in\left\{0;-2\right\}\)
\(Vậy\)\(n\in\left\{0;-2\right\}\)
\(b.n-3⋮n+1\Leftrightarrow n+1-4⋮n+1\Leftrightarrow4⋮n+1\)\(\left(vìn+1⋮n+1\right)\)
\(\Leftrightarrow n+1\inƯ\left(\text{4}\right)\Leftrightarrow n+1\in\left\{\pm1;\pm2;\pm4\right\}\)\(\Leftrightarrow n\in\left\{0;-2;1;-3;3;-5\right\}\)
\(Vậy\)\(n\in\left\{0;-2;1;-3;3;-5\right\}\)
