\(\frac{a^2-3a-5}{a-2}\left(1\right)=\frac{a\left(a-2\right)-\left(a+5\right)}{a-2}\)
\(=a-\frac{a+5}{a-2}=a-\frac{a-2+7}{a-2}\)
\(=a-1+\frac{7}{a+2}\)
để (1) thuộc Z thì 7 phải chia hết cho a+2
\(\Rightarrow a+2\inƯ\left(7\right)=\left\{\pm1;\pm7\right\}\)
=> a={-1;-3;5;-9}
Ta có \(\frac{a^2-3a-5}{a-2}=\frac{a^2-2a-a+2-7}{a-2}=\frac{a\left(a-2\right)-\left(a-2\right)-7}{a-2}=\frac{\left(a-2\right)\left(a-1\right)-7}{a-2}\)
\(=a-1-\frac{7}{a-2}\)
Vì \(\hept{\begin{cases}a\inℤ\\-1\inℤ\end{cases}}\Rightarrow\frac{-7}{a-2}\inℤ\Rightarrow-7⋮a-2\Rightarrow a-2\inƯ\left(-7\right)\)
=> \(a-2\in\left\{1;7;-1;-7\right\}\)
=> \(a\in\left\{3;9;1;-5\right\}\)
Vậy \(a\in\left\{3;9;1;-5\right\}\)l là giá trị cần tìm
