a) 3x(x - 1) - 1 + x = 0
  3x(x - 1) + (x - 1) = 0
       (x - 1)(3x + 1) = 0
   TH1) x - 1 = 0
                 x = 0 +1
                 x = 1
   TH2) 3x - 1 = 0
                 3x = 0 +1
                   x = 1: 3
                   x =\(\dfrac{1}{3}\)
Vậy x ϵ (1; \(\dfrac{1}{3}\))
b) x² - 9x = 0
     x(x-9) = 0
TH1) x = 0
TH2) x - 9 = 0
              x = 0 + 9
              x = 9
Vậy x ϵ (0; 9)

a) x² + 25 - 10x
= x² - 2.x.5 + 5²
= (x - 5)²
b) -8y³ + x³
= x³ - (2y)³
= (x - 2y)(x² + 2xy + 4y²⁾

a) (2x + 1)²
= (2x)² + 2.2x.1 + 1²
= 4x² + 2x + 1
b) (a - \(\dfrac{b}{2}\))3
= a³ - 3.a².\(\dfrac{b}{2}\)+ 3.a.(\(\dfrac{b}{2}\))² - \(\dfrac{b}{2}\)³

a) x² + 2xy + y² - x - y
= (x² + 2.x.y + y²) - (x - y)
= (x + y)² - ( x - y)
= (x + y)(x + y - 1)
b) 2x³ + 6x² + 12x + 8