a) $ĐKXĐ : \left\{ \begin{array}{l}2x-2 \neq 0\\x^2-1 \neq 0 \\ 2x+2 \neq 0 \end{array} \right.$
$⇔ \left\{ \begin{array}{l}x \neq 1\\x \neq ±1 \\ x \neq -1 \end{array} \right.$
$⇔ \left\{ \begin{array}{l}x \neq 1 \\ x \neq -1 \end{array} \right.$
Vậy $ĐKXĐ$ để biểu thức xác định là $x \neq 1, x \neq -1$
b) Ta có :
$A = \bigg(\dfrac{x+1}{2x-2}+\dfrac{3}{x^2-1} - \dfrac{x+3}{2x+2} \bigg) . \dfrac{4x^2-4}{5}$
$=\dfrac{(x+1)^2+3.2-(x+3).(x-1)}{2.(x-1)(x+1)}. \dfrac{4(x^2-1)}{5}$
$=\dfrac{x^2+2x+1+6-x^2-2x+3}{2(x-1)(x+1)}.\dfrac{4(x-1)(x+1)}{5} $
$=\dfrac{10}{2.(x-1)(x+1)}.\dfrac{4.(x-1)(x+1)}{5} $
$=4$
