\(\frac{2a+5}{a+1}=\frac{2\left(a+1\right)+3}{a+1}=\frac{2\left(a+1\right)}{a+1}+\frac{3}{a+1}=2+\frac{3}{a+1}\in Z\)
\(\Rightarrow3⋮a+1\)
\(\Rightarrow a+1\inƯ\left(3\right)=\left\{1;3;-1;-3\right\}\)
\(\Rightarrow a\in\left\{0;2\right\}\left(a\in N\right)\)
Ta có:\(2a+5⋮a+1\)
\(\Leftrightarrow2a+2+3⋮a+1\)
\(\Leftrightarrow2\left(a+1\right)+3⋮a+1\)
\(\Leftrightarrow3⋮a+1\)
\(\Leftrightarrow a+1\inƯ\left(3\right)\)
Mà \(a\in N\Rightarrow a+1\ge1\)
\(\Leftrightarrow a+1\in\left\{1,3\right\}\)
\(\Leftrightarrow a\in\left\{0,2\right\}\)
Vậy a=0 hoặc a=2
Ta có: \(2a+5⋮a+1\)
\(\Rightarrow\left(2a+2\right)+3⋮a+1\)
\(\Rightarrow2\left(a+1\right)+3⋮a+1\)
\(\Rightarrow3⋮a+1\)
\(\Rightarrow a+1\in\left\{1;-1;3;-3\right\}\)
Mà \(a\in N\Rightarrow a+1\in\left\{1;3\right\}\)
+) \(a+1=1\Rightarrow a=0\)
+) \(a+1=3\Rightarrow a=2\)
Vậy \(a\in\left\{0;2\right\}\)