Réponse :
Explications étape par étape
Bonjour
(2x - 16)(5x + 14) << 0
2(x - 8)(5x + 14) << 0
x - 8 = 0 et 5x + 14 = 0
x = 8 et 5x = -14
x = 8 et x = -14/5
x............|-inf.......-14/5.......8.............+inf
x - 8......|......(-).............(-)...o...(+)........
5x + 14.|......(-)......o.....(+)........(+)....,,,,
Ineq......|......(+).....o.....(-)...o....(+).......
[tex]x \in [-14/5 ; 8][/tex]
(3 - x)/(8x + 48)(-2x + 11) << 0
(3 - x)/[8(x + 6)(-2x + 11)] << 0
x + 6 # 0 et -2x + 11 # 0
x # -6 et 2x # 11
x # -6 et x # 11/2
3 - x = 0
x = 3
x............|-inf...........-6...........3..........11/2.......+inf
3 - x......|........(+)...........(+).....o....(-)..........(-)........
x + 6.....|........(-)......o....(+)..........(+)..........(+).......
-2x + 11.|........(+)...........(+)...........(+)...o....(-)........
Ineq.....|.........(-).....||.....(+)....o.....(-)....||.....(+)........
[tex]x \in ]-\infty ; -6[ U [3 ; 11/2[[/tex]
(3 - x)(8x + 48)/(-2x + 11) << 0
(3 - x) * 8(x + 6) / (-2x + 11) << 0
3 - x = 0 ou x + 6 = 0 et -2x + 11 # 0
x = 3 ou x = -6 et 2x # 11
x = 3 ou x = -6 et x # 11/2
x............|-inf...........-6...........3..........11/2.......+inf
3 - x......|........(+)...........(+).....o....(-)..........(-)........
x + 6.....|........(-)......o....(+)..........(+)..........(+).......
-2x + 11.|........(+)...........(+)...........(+)...o....(-)........
Ineq.....|.........(-).....o.....(+)....o.....(-)....||.....(+)........
[tex]x \in [3 ; 11/2[[/tex]
