Aidez moi svp merciii d'avance

Réponse :

Bonjour

Explications étape par étape

2 - 3x = 0

3x = 2

x = 2/3

2x^2 + 3 toujours positif quelque soit x

x..............|-inf..................2/3...............+inf

2-3x.......|...........(+)...........o.......(-).............

2x^2+3..|...........(+).....................(+)...........

Ineq.......|............(+)...........o.......(-)..............

S = [2/3 ; +inf [

x^2 < 4

x^2 - 4 < 0

(x - 2)(x + 2) < 0

x - 2 = 0 => x = 2

x + 2 = 0 => x = -2

x.............| -inf.............(-2).............(2)........+inf

x - 2.......|...........(-)................(-).......o....(+).......

x + 2......|...........(-).......o.......(+)............(+).......

Ineq......|............(+)......||........(-)......||......(+)......

S = ] -2 ; 2 [

(4x - 3)^2 << 0

Pas de solution car un carré est toujours positif

-2(x + 5)(2x - 7) < 0

x + 5 = 0 => x = -5

2x - 7 = 0 => 2x = 7 => x = 7/2

x...........| -inf..........(-5)..........(7/2)........+inf

x + 5....|.........(-)......o.....(+)............(+).........

2x - 7..|.........(-)..............(-)......o.....(+).........

Ineq....|.........(+)......||.....(-)......||......(+).........

S = ] -5 ; 7/2 [