\(x=\frac{3}{4a+1}\)
Ta có U(3)={1;3;-1;-3}
mà \(x\in\)N*
=>x={1;3}
TH1: 4a+1 =1
4a=1-1
4a=0
a=0:4
a=0
TH2: 4a+1 =3
4a=3-1
4a=2
a=2:4
a=\(\frac{2}{4}\)
a=\(\frac{1}{2}\)
Vậy a={0;\(\frac{1}{2}\)}