\(a,A=\dfrac{x^2+x-2-x^2-x}{\left(x+1\right)\left(x-1\right)}\cdot\dfrac{3\left(x+1\right)}{2}=\dfrac{-6}{2\left(x-1\right)}=\dfrac{3}{1-x}\\ b,A< 0\Leftrightarrow1-x< 0\left(3>0\right)\Leftrightarrow x>1\\ c,A\in Z\Leftrightarrow1-x\inƯ\left(3\right)=\left\{-3;-1;1;3\right\}\\ \Leftrightarrow x\in\left\{-2;0;2;4\right\}\)


