Bonjour,

Ex1

A = (3/4 + 1/5) : 2/3 = [(3 × 5 + 1 × 4) / (4 × 5)] × [3/2]

A = 19/20 × 3 / 2 = (19 × 3) / (20 × 2) = 57/40

B = 5/7 - 3/7 × 2/5 = 5/7 - 6/35 = (5 × 5) / (7 × 5) - 6/35

B = 25/35 - 6/35 = 19/35

Ex2

1) A = (2x - 3)² - (2x - 3) (x + 1)

A = 4x² - 12x + 9 - 2x² - 2x + 3x + 3

A = 2x² - 11x + 12

2) A = (2x - 3)² - (2x - 3) (x + 1)

A = (2x - 3) ((2x - 3) - (x + 1))

A = (2x - 3) (2x - 3 - x - 1)

A = (2x - 3) ( x - 4)

3) (2x - 3) (x - 4) = 0 équivaut à 2x - 3 = 0 ou x - 4 = 0

Soit x = 3/2 ou x = 4

L'ensemble des solution est donc S = {3/2 ; 4}

Ex3

A = 1 - 1/(x+2) = (x + 2) / (x + 2) - 1 / (x + 2) = (x + 2 - 1)/(x + 2) = (x + 1) / (x + 2)

B = 1 / (x - 1) - x / (x + 1) = [(x + 1) - x (x - 1)] / [(x - 1) (x + 1)]

B = (x + 1 - x² + x) / (x² - 1) = (-x² + 2x + 1) / (x² - 1)