Réponse :

Bonjour

x²(x - 1) - 2x(x - 1) < 1 -x

⇔ x²(x - 1) - 2x(x - 1) - (1 - x) < 0

⇔ x²(x - 1) - 2x(x - 1) + (x - 1) < 0

⇔ (x - 1)(x² - 2x - 1) < 0

⇔ (x - 1)(x - 1)² < 0

⇔ x - 1 < 0

⇔ x < 1

S = ]-∞ ; 1[