Bonjour,
a) (3x+1)²-(3x-1)² = (9x²+6x+1)-(9x²-6x+1)
= 9x²-9x²+6x+6x+1-1
= 12x
b) (3x+1)²-(3x-1)² = [(3x+1)+(3x-1)][(3x+1)-(3x-1)]
= (6x) × (2)
c) posons x = 1 000
on a alors alors 3 001² = (3x + 1)²
et 2 999 = (3x - 1)²
comme on sait que (3x+1)²-(3x-1)² = 12x
alors 3 001² - 2 999² = 12 × 1 000 = 12 000
d) D = 0 ⇒ (3x+1)²-(3x-1)² = 0
⇒ 12x = 0
⇒ x = 0
