a) ( x + 3 )( x2 - 3x + 9 ) - x( x - 2 )2 = 27
⇔ x3 + 27 - x( x2 - 4x + 4 ) = 27
⇔ x3 + 27 - x3 + 4x2 - 4x = 27
⇔ 4x2 - 4x + 27 - 27 = 0
⇔ 4x2 - 4x = 0
⇔ 4x( x - 1 ) = 0
⇔ 4x = 0 hoặc x - 1 = 0
⇔ x = 0 hoặc x = 1
b) ( x - 1 )( x - 5 ) + 3 = 0
⇔ x2 - 5x - x + 6 + 3 = 0
⇔ x2 - 6x + 9 = 0
⇔ ( x - 3 )2 = 0
⇔ x - 3 = 0
⇔ x = 3