Bonjour
Identité remarquable à connaître :
(a + b)² = a² + 2ab + b²
(a - b)² = a² - 2ab + b²
a² - b² = (a + b)(a - b)
Quand je marque IR, cela signifie que j'utilise une identité remarquable.
Exercice 2
1. A = (x - 4) (5x + 1) + x - 4
= (x*5x + x*1 - 4*5x - 4*1) + x - 4
= (5x² + x - 20x - 4) + x - 4
= 5x² - 19x - 4 + x - 4
= 5x² - 18x - 8
B = (4x - 1) (3x + 7) + (3x + 7)² => IR
= (4x*3x + 4x*7 - 1*3x - 1*7) + [(3x)² + 2*3x*7 + 7²]
= 12x² + 28x - 3x - 7) + (9x² + 42x + 49)
= 12x² + 25x - 7 + 9x² + 42x + 49
= 21x² + 67x + 42
D = (2x+1)² - (7x-5)² => IR
= [(2x)² + 2*2x*1 + 1²] - [(7x)² - 2*7x*5 + 5²]
= (4x² + 4x + 1) - (49x² - 70x + 25)
= 4x² + 4x + 1 - 49x² + 70x - 25
= -45x² + 74x - 24
2. A = (x - 4) (5x + 1) + x - 4
= (x - 4) (5x + 1) + (x - 4) x 1
= (x - 4)[(5x + 1) + 1]
= (x - 4)(5x + 1 + 1)
= (x - 4)(5x + 2)
B = (4x - 1) (3x + 7) + (3x + 7)² => IR
= (3x + 7)[(4x - 1) + (3x + 7)]
= (3x + 7)(4x - 1 + 3x + 7)
= (3x + 7)(7x + 6)
C = 121x² + 44x + 4 => IR
= (11x)² + 2*11*2 + 2²
= (11x + 2)²
D = (2x+1)² - (7x-5)² => IR
= [(2x + 1) + (7x - 5)][(2x + 1) - (7x - 5)]
= (2x + 1 + 7x - 5)(2x + 1 - 7x - 5)
= (9x - 4)(-5x - 4)
Exercice 3
A = (7 + x)(5x - 6) + 5x - 6
= (7 + x)(5x - 6) + (5x - 6) x 1
= (5x - 6)[(7 + x) + 1]
= (5x - 6)(7 + x + 1)
= (5x - 6)(x + 8)
B = (2x - 5)(x + 15) - 2x + 5
= (2x - 5)(x + 15) - (2x - 5) x 1
= (2x - 5)[(x + 15) - 1]
= (2x - 5)(x + 15 - 1)
= (2x - 5)(x + 14)
Voilà j'espère avoir pu t'aider ^^