(x - 3)(x - 2)(x + 1)(x + 2) = 12
=> (x - 3)(x + 2)(x - 2)(x + 1) = 12
=> (x2 - x - 6)(x2 - x - 2) = 12
Đặt a = x2 - x - 2
=> (a - 4)a = 12
=> a2 - 4a = 12
=> a2 - 4a + 4 = 16
=> (a - 2)2 - 16 = 0
=> (a - 2 - 16)(a - 2 + 16) = 0
=> (x2 - x - 20)(x2 - x + 12) = 0
=> x2 - x - 20 = 0 hoặc x2 - x + 12 = 0
=> x(x - 1) = 20 => x(x - 1) = - 12
=> x = 5 => x = - 4 hoặc x = 4