Ta có \(t=\frac{3x-8}{x-5}=\frac{3x-15+7}{x-5}=\frac{3\left(x-5\right)}{x-5}+\frac{7}{x-5}=3+\frac{7}{x-5}\)
Để t là số nguyên khi và chỉ khi \(\frac{7}{x-5}\)nguyên
\(\Rightarrow\left(x-5\right)\in\text{Ư}\left(7\right)=\left\{-7;-1;1;7\right\}\)
\(\cdot x-5=-7\Leftrightarrow x=-2\left(tm\right)\)
\(\cdot x-5=-1\Rightarrow x=4\left(tm\right)\)
\(x-5=1\Rightarrow x=6\left(tm\right)\)
\(\cdot x-5=7\Rightarrow x=12\left(tm\right)\)
Vậy \(x\in\left\{-2;4;6;12\right\}\) thì t nguyên