Giải:
Để \(\frac{3a+10}{a+6}\in Z\Rightarrow3a+10⋮a+6\)
Ta có: \(3a+10⋮a+6\)
\(\Rightarrow\left(3a+18\right)-8⋮a+6\)
\(\Rightarrow3\left(a+6\right)-8⋮a+6\)
\(\Rightarrow8⋮a+6\)
\(\Rightarrow a+6\in\left\{1;-1;2;-2;4;-4;8;-8\right\}\)
\(\Rightarrow a\in\left\{-5;-7;-4;-8;-2;-10;2;-14\right\}\)
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