ta có:
\(\dfrac{2a+1}{a^2+3a-1}=\dfrac{2\left(a+1\right)-1}{a^2+2a+1+a-2}\)
=\(\dfrac{2\left(a+1\right)-1}{\left(a+1\right)^2+a-2}\)
=\(\dfrac{2-1}{a+1+a-2}\)
= \(\dfrac{1}{2a-1}\)
\(\dfrac{2a+1}{a^2+3a-1}\in Z\Leftrightarrow\dfrac{1}{2a-1}\in Z\)
\(\Leftrightarrow1⋮2a-1\)
=> 2a - 1\(\inƯ_{\left(1\right)}=\left\{\pm1\right\}\)
=> 2a - 1 = {-1;1}
=> 2a={0;2}
=> a={0;1}
vậy.....