bonjour je dois résoudre ces équations [tex]a) \: 4(2 - x) + 5 = - 3(2x + 3) - 12 \\ b) \: \frac{3x + 1}{4} = \frac{5x - 2}{5} \\ c) \: \frac{x}{3} + \

Bonsoir,

a) 4(2-x)+5 = -3(2x+3)-12
8-4x+5 = -6x-9-12
13-4x = -6x-21
13-4x-13+6x = -6x-21-13+6x
2x = -34
x = -34/2 = -17

b) ((3x+1)/4) = ((5x-2)/5)
(5(3x+1)/20) = (4(5x-2)/20)
5(3x+1) = 4(5x-2)
15x+5 = 20x-8
20x-8 = 15x+5
20x-8-15x+8 = 15x+5-15x+8
5x = 13
x = 13/5

c) (x/3)+(5/6) = (1/2)-x
(x/3)+(5/6)-(5/6)+x = (1/2)-x-(5/6)+x
(x/3)+x = (1/2)-(5/6)
(x/3)+(3x/3) = (3/6)-(5/6)
4x/3 = -2/6
4x/3 = -1/3
4x = -1
x = -1/4

d) x-(x+1) = (x+3)-(x-3)
x-x-1 = x+3-x+3
-1 = 6
Donc cette équation n'admet aucune solution dans l'ensemble des nombres réels.