làm giúp mik vs ah..

a

Ta có

$A = \dfrac{8}{(x^2+3)(x^2-1)} + \dfrac{2}{x^2+3} + \dfrac{1}{x+1}$
ĐKXĐ: $x \ne \pm 1$
$A = \dfrac{8 + 2(x^2-1) + (x^2+3)(x-1)}{(x^2+3)(x-1)(x+1)}$
$A = \dfrac{8 + 2x^2-2 + x^3-x^2+3x-3}{(x^2+3)(x^2-1)}$
$A = \dfrac{x^3+x^2+3x+3}{(x^2+3)(x^2-1)} = \dfrac{x^2(x+1)+3(x+1)}{(x^2+3)(x^2-1)}$
$A = \dfrac{(x^2+3)(x+1)}{(x^2+3)(x^2-1)} = \dfrac{(x^2+3)(x+1)}{(x^2+3)(x-1)(x+1)}$
$A = \dfrac{1}{x-1}$
b

Ta có

$B = \dfrac{x^3+x^2-2x-20}{x^2-4} - \dfrac{5}{x+2} + \dfrac{3}{x-2}$.
ĐKXĐ: $x \ne \pm 2$.
$B = \dfrac{x^3+x^2-2x-20 - 5(x-2) + 3(x+2)}{(x-2)(x+2)}$
$B = \dfrac{x^3+x^2-2x-20-5x+10+3x+6}{(x-2)(x+2)}$
$B = \dfrac{x^3+x^2-4x-4}{(x-2)(x+2)} = \dfrac{x^2(x+1)-4(x+1)}{(x-2)(x+2)}$
$B = \dfrac{(x^2-4)(x+1)}{(x-2)(x+2)} = \dfrac{(x-2)(x+2)(x+1)}{(x-2)(x+2)}$
$B = x+1$