$a.A=\dfrac{x^3+2x^2+x}{x^3-x}=\dfrac{x(x^2+2x+1)}{x(x^2-1)}=\dfrac{(x+1)^2}{(x+1)(x-1)}=\dfrac{x+1}{x-1} \\Để\ A= 2 \\⇒\dfrac{x+1}{x-1}=2 \\⇔2(x-1)=x+1 \\⇔2x-2=x+1 \\⇔x=3 \\b.Ta\ có : \\A=\dfrac{x+1}{x-1}=\dfrac{x-1+2}{x-1}=1+\dfrac{2}{x-1}$
Để A có giá trị nguyên
$⇒x-1∈Ư(2)=\{±1;±2\} \\+)x-1=1⇔x=2 \\+)x-1=-1⇔x=0 \\+)x-1=2⇔x=3 \\+)x-1=-2⇔x=-1$