\(B=\frac{3x+4}{x-3}\inℤ\left(x\ne3\right)\)
\(\Rightarrow3x+4⋮x-3\)
\(\Rightarrow3x-9+13⋮x-3\)
\(\Rightarrow3\left(x-3\right)+13⋮x-3\)
Ta có: \(3\left(x-3\right)⋮x-3\)
\(\Rightarrow13⋮x-3\)
\(\Rightarrow x-3\inƯ\left(13\right)=\left\{\pm1;\pm13\right\}\Rightarrow x\in\left\{4;2;16;-10\right\}\)