Để \(x=\frac{5}{2a-1}\) là số nguyên thì \(5⋮2a-1\)
\(\Rightarrow2a-1\inƯ\left(5\right)\)
Mà \(Ư\left(5\right)=\left\{1;-1;5;-5\right\}\)
\(\Rightarrow2a-1\in\left\{1;-1;5;-5\right\}\)
Ta lập bảng sau:
2a-1 | 1 | -1 | 5 | -5 |
a | 1 | 0 | 3 | -2 |
ĐK \(a\in Z\) | TM | TM | TM | TM |
Vậy \(a\in\left\{1;0;3;-2\right\}\).
x nguyên khi \(2a-1\inƯ\left(5\right)=\left\{-5;-1;1;5\right\}\)
\(2a-1=-5\Rightarrow a=-2\)\(2a-1=-1\Rightarrow a=0\)\(2a-1=1\Rightarrow a=1\)\(2a-1=5\Rightarrow a=3\)