Réponse :
Explications étape par étape
Bonjour
Résoudre les inéquations :
2x > x^3
x^3 - 2x < 0
x(x^2 - 2) < 0
x(x - V2)(x + V2) < 0 (avec V : racine de)
x........|-inf.............(-V2)......0........V2........+inf
x........|.........(-)................(-)..o...(+).........(+)........
x-V2.|..........(-)................(-).......(-)...o.....(+).......
x+V2|..........(-).........o.....(+)......(+)..........(+)......
Ineq.|...........(-)........o.....(+).o...(-)....o....(+)......
[tex]x \in ]-\infty ; -\sqrt{2}[ U ]0 ; \sqrt{2}[[/tex]
(x + 2)(6x - 1) < (x + 2)(x - 3)
(x + 2)(6x - 1) - (x + 2)(x - 3) < 0
(x + 2)(6x - 1 - x + 3) < 0
(x + 2)(5x + 2) < 0
x + 2 = 0 ou 5x + 2 = 0
x = -2 ou x = -2/5
x..........|-inf.........(-2)...........(-2/5).........+inf
x + 2...|........(-).....o.....(+)...............(+)..........
5x + 2|.........(-).............(-)........o.....(+).........
Ineq...|..........(+)...o......(-).........o.....(+)........
[tex]x \in ]-2 ; -2/5[[/tex]
(x + 1)^2 << 9
(x + 1)^2 - 3^2 << 0
(x + 1 - 3)(x + 1 + 3) << 0
(x - 2)(x + 4) << 0
x - 2 = 0 ou x + 4 = 0
x = 2 ou x = -4
x............|-inf..........(-4)..........2..........+inf
x - 2......|........(-).............(-).....o....(+)........
x + 4.....|.........(-)......o....(+)...........(+)........
Ineq......|........(+)......o.....(-).....o....(+).......
[tex]x \in [-4 ; 2][/tex]
(2x + 3)^2 - (-5x + 6)^2 < 0
(2x + 3 + 5x - 6)(2x + 3 - 5x + 6) < 0
(7x - 3)(-3x + 9) < 0
(7x - 3) * 3(-x + 3) < 0
3(7x - 3)(-x + 3) < 0
7x - 3 = 0 ou -x + 3 = 0
7x = 3 ou x = 3
x = 3/7 ou x = 3
x..............|-inf............3/7.............3.........+inf
7x - 3......|........(-)........o......(+)..........(+)........
-x + 3......|........(+).................(+)....o...(-)........
Ineq.......|.........(-)........o.......(+)...o.....(-).......
[tex]x \in ]-\infty ; 3/7[ U ]3 ; +\infty[[/tex]
