Để \(x\inℤ\) thì \(\frac{a-5}{a}\inℤ\)
Ta có: \(\frac{a-5}{a}=\frac{a}{a}-\frac{5}{a}=1-\frac{5}{a}\)
Để \(\frac{a-5}{a}\inℤ\) thì \(\frac{5}{a}\inℤ\)
\(\Rightarrow a\inƯ\left(5\right)=\left\{\pm1;\pm5\right\}\)
Trả lời:
Ta có: \(x=\frac{a-5}{a}\)\(=\frac{a}{a}-\frac{5}{a}=1-\frac{5}{a}\)
Để \(x\inℤ\)thì \(\frac{5}{a}\inℤ\)
\(\Rightarrow5⋮a\)hay \(a\inƯ\left(5\right)=\left\{\pm1;\pm5\right\}\)
Vậy \(a\in\left\{\pm1;\pm5\right\}\)thì \(x\inℤ\)