\(A=\frac{2x+1}{x-2}=\frac{2x-4+5}{x-2}=2+\frac{5}{x-2}\)
Để A thuộc Z thì 5/(x-2) thuộc Z hay \(x-2\inƯ\left(5\right)=\left\{\pm1;\pm5\right\}\)
\(\Leftrightarrow x\in\left\{3;1;7;-3\right\}\)
\(B=\frac{-x+8}{x+1}=\frac{-\left(x+1\right)-7}{x+1}=-1+\frac{-7}{x+1}\)
Để \(A\inℤ\Leftrightarrow\frac{-7}{x+1}\Leftrightarrow x+1\inƯ\left(-7\right)=\left\{\pm1;\pm7\right\}\)
\(\Leftrightarrow x\in\left\{0;-2;6;-8\right\}\)