a,
$\text{4x (x + 1) = 8 (x + 1)}$
$\text{⇔ 4x (x + 1) - 8 (x +1)= 0}$
$\text{⇔ (x+1) (4x-8) =0}$
$\text{⇔ x+1 = 0 hoặc 4x-8 = 0}$
$\text{⇒ x= -1 hoặc 2}$
$\text{⇒ x ∈ {-1 ; 2}}$
b,
$\text{x³ = x^5}$
$\text{x^5 - x³ = 0}$
$\text{x³ . (x² - 1) = 0}$
$\text{x² - 1 = 0 hoặc x³ = 0}$
$\text{⇒ x = 0 hoặc ± 1}$