Đáp án:
$=\dfrac{3(x-2)}{x+1}$
Giải thích các bước giải:
$\begin{array}{l}A=\dfrac{3x-3}{x+2}.\dfrac{x^2-4}{x^2-1}\\ĐKXĐ:x \neq 2,-2,1,-2\\A=\dfrac{3(x-1)}{x+2}.\dfrac{(x-2)(x+2)}{(x-1)(x+1)}\\=\dfrac{3(x-1)(x-2)(x+2)}{(x+2)(x-1)(x+1)}\\=\dfrac{3(x-2)}{x+1}\\\end{array}$