a/ $\dfrac{x^2+x+1}{-2x^2+2x-2}=\dfrac{x^2+x+x}{-2(x^2-x+1)}=\dfrac{1}{-2}<0$
b/ $\dfrac{x^2+2x+1}{3x^2+2x-5}=\dfrac{(x+1)^2}{3x^2+5x-3x-5}$
$=\dfrac{(x+1)^2}{3x(x-1)+5(x-1)}$
$=\dfrac{(x+1)^2}{(x-1)(3x+5)}$
$A\in \mathbb{Z}$
$\leftrightarrow \begin{cases}x+1>0\\x-1>0\\3x+5>0\end{cases}$
$\leftrightarrow \begin{cases}x>-1\\x>1\\3x>-5\end{cases}$
$\leftrightarrow \begin{cases}x>-1\\x>1\\x>-\dfrac{5}{3}\end{cases}$