faudrait savoir appliquer :

k (a+b) = ka + kb

et (a+b) (c+d) = ac + ad + bc + bd

et penser à a² - b² = (a+b) (a-b)

3x (2 + x ) = 3x*2 + 3x*x = 3x² + 6x

(4+x) (x-7) = 4*x + 4*(-7) + x*x + x*(-7) = 4x - 28 + x² - 7x = x² - 3x - 28

(x-5)² comme (a-b)² = a² - 2ab +b² tu auras x² - 10x + 25

12² + 5x = x (12x + 5)

1/2x² - 1/4x = x (1/2x - 1/4)

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

3/5x - 2/5x = -1 - 5

1/5x = -6

x = -6*5 = -30

(x+12) (3x-4) = 0

soit x + 12 = 0 => x = -12

soit 3x -4= 0 => x = 4/3

(x+2)² - 9 = 0

(x+2)² - 3² = 0

(x+2+3) (x+2-3) = 0

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

soit x = -5 soit x = 1