Réponses

  • Omnes
  • Modérateur confirmé
2014-03-03T22:23:56+01:00
Salut,

1) 3x²+2x+7<9x+7
3x² + 2x + 7 -9x - 7 < 0
3x² -7x < 0
x(3x - 7) < 0
x < 0
3x - 7 < 0
3x < 7
x < 7/3

S=]-infini; 7/3[

2) 4x-1/3-x≥5


3x - 1/3 ≥ 5
3x ≥ 5 + 1/3
3x ≥ 16/3
x ≥ (16/3) / 3
x ≥ 16/9

S = [16/9; + infini[


Bonne soirée !

2014-03-03T22:53:06+01:00
1) 3x² + 2x + 7 < 9x + 7
3x² + 2x + 7 -9x - 7 < 0
3x² -7x < 0
x (3x - 7) < 0
x < 0
3x - 7 < 0
3x < +7
x < 7/3

Solution = ] -Infini ; 7/3 [

2) 4x - 1/3 - x ≥ 5

3x - 1/3 ≥ 5
3x ≥ 5 +1/3
3x ≥ 16/3
x ≥ (16/3)/3
x ≥ 16/9

Solution = [ 16/9 ; + Infini [