b) (2x – 1)² – (x + 3)² = 0

⇔ (2x – 1 – x – 3)(2x – 1 + x + 3) = 0

⇔ (x – 4)(3x + 2) = 0

$\iff \left[\begin{array}{l}x-4=0\\ 3\text{x}+2=0\end{array}\right.$$\iff \left[\begin{array}{l}x=4\\ x=\frac{-2}{3}\end{array}\right.$

Vậy $x\in \left\{4;\frac{-2}{3}\right\}$  .

c) x²(x – 3) + 12 – 4x = 0

⇔ x²(x – 3) – 4(x – 3) = 0

⇔ (x – 3)(x² – 4) = 0

⇔ (x – 3)(x – 2)(x + 2) = 0 

$\iff \left[\begin{array}{c}x-3=0\\ x-2=0\\ x+2=0\end{array}\iff \left[\begin{array}{c}x=3\\ x=2\\ x=-2\end{array}\right.\right.$

Vậy x ∈ {3; 2; –2}.