a) Ta có : $x^3 + 8 = x^3 + 2^3$
$ = (x+2).(x^2-2x+4)$
Do đó : $\dfrac{x^3+8}{(1-x).(x^2-2x+4)} = \dfrac{(x+2).(x^2-2x+4)}{(1-x).(x^2-2x+4)}$
$ = \dfrac{x+2}{1-x}$
$ = \dfrac{-x-2}{x-1}$ với $ĐKXĐ : x \neq 1$
b) Ta có : $x^2+5x+4$
$ = x^2+x+4x+4=(x+1).(x+4)$
$2x^2-x-3 $
$ = 2x^2+2x-3x-3 = (x+1).(2x-3)$
$x^2+3x-4 $
$ = x^2-x+4x-4 = (x-1).(x+4)$
$2x^2-5x+3$
$ = 2x^2-2x-3x+3=(x-1).(2x-3)$
Do đó : $\dfrac{x^2+5x+4}{2x^2-x-3} = \dfrac{(x+1).(x+4)}{(x+1).(2x-3)} = \dfrac{x+4}{2x-3}$ với $x \neq -4, \dfrac{3}{2}$
$\dfrac{x^2+3x-4}{2x^2-5x+3} = \dfrac{(x-1).(x+4)}{(x-1).(2x-3)} = \dfrac{x+4}{2x-3}$ với $x \neq 1, \dfrac{3}{2}$
Vì vậy $\dfrac{x^2+5x+4}{2x^2-x-3} = \dfrac{x^2+3x-4}{2x^2-5x+3}$
