$a/A=\dfrac{x-1}{x-2}-\dfrac{3}{x^2-x-2}+\dfrac{1}{x+1}$
$=\dfrac{x-1}{x-2}-\dfrac{3}{(x-2)(x+1)}+\dfrac{1}{x+1}$
$=\dfrac{(x-1)(x+1)}{(x-2)(x+1)}-\dfrac{3}{(x-2)(x+1)}+\dfrac{x-2}{(x-2)(x+1)}$
$=\dfrac{x²-1-3+x-2}{(x-2)(x+1)}$
$=\dfrac{x^2+x-6}{(x-2)(x+1)}$
$=\dfrac{(x-2)(x+3)}{(x-2)(x+1)}$
$=\dfrac{x+3}{x+1}$
b/Để A > 0
⇒ \(\left[ \begin{array}{l}\left\{\begin{matrix} x+3>0 & \\ x+1>0 & \end{matrix}\right.\\\left\{\begin{matrix} x+3<0 & \\ x+1<0 & \end{matrix}\right.\end{array} \right.\) ⇒\(\left[ \begin{array}{l}x>-2\\x<-3\end{array} \right.\)
c/Ta có :
$\dfrac{x+3}{x+1}=\dfrac{x+1}{x+1}+\dfrac{2}{x+1}=1+\dfrac{2}{x+1}$
Để A là số nguyên dương thì 2 phải chia hết cho x+1
$⇒x+1∈Ư(2)=\{±1;±2\}$
$⇒x∈\{0;-2;1;-3\}$
