bonjour pouvez vous m'aider a resoudre ces equations merci : (7x - 5 ) au carre = 16 10(x + 7 ) (x - 5 ) = 3x (x+7) x au carre = 81 (7x - 5 ) au carre = (2x + 1

bjr

(7x - 5 ) au carre = 16

(7x-5)² - 4² = 0

(7x - 5 + 4) (7x - 5 - 4) = 0

(7x - 1) (7x - 9) = 0

soit x = 1/7 soit x = 9/7

10(x + 7 ) (x - 5 ) = 3x (x+7)

(x+7) (10(x-5) - 3x) = 0

(x+4) (10x-50-3x) = 0

(x+4) (7x - 50)=0

x = -4 ou x = 50/7

x² = 81

x² - 81 = 0

(x+9) (x-9) = 0

soit x = -9 soit x = 9

(7x - 5 )² - (2x + 1)² = 0

(7x-5 + 2x+1) (7x-5 - (2x+1) = 0

tu finis

pour la premiere :

(7x - 5 ) au carre = 16

(7x-5)² - 4² = 0

(7x - 5 + 4) (7x - 5 - 4) = 0

(7x - 1) (7x - 9) = 0

soit x = 1/7 soit x = 9/7