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wendy14
@wendy14
January 2021
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bonsoir je voudrai de l'aide pour l'exercice de math . merci d'avance
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anylor
Bonsoir
3.1)
|x+7|≥3
1er cas
x+7 ≥ 0 => x+7 ≥3 => x ≥ 3-7 => x ≥ - 4
2nd cas
x+7 ≤ 0 => - (x +7 ) ≥ 3 => -x -7≥3 => x ≤ -7-3 => x ≤ -10
solution :
x≤ -10 OU x ≥ -4
3.2)
|3x-1| ≤5/2
1er cas
3x-1≥0 => 3x-1 ≤ 5/2 => 3x ≤ 5/2 +2/2 => 3x ≤ 7/2
x ≤ 7/6
2nd cas
3x-1≤0 => -(3x-1) ≤ 5/2 => -3x +1 ≤ 5/2
-3x ≤ 5/2 -2/2
-3x ≤ 3/2
3x ≥ -3/2
x ≥ - 3/6 => x ≥ -1/2 ( on simplifie 3/6 = 1/2)
solution
-1/2 ≤ x ≤ 7/6
3.3)
|x+6| ≤0
1er cas
x+6 ≤0 => x ≤ -6
2nd cas
-x-6 ≤0 => x ≥-6
1 seule solution possible x= - 6
3. 4)
|x| +x = 0
1er cas
x>0
x+x = 0 => 2x =0 => x =0
2nd cas
si x <0
-x + x = 0 0 = 0 toujours vrai
donc solution :
x ≤ 0
3.5)
il faut faire un tableau avec les différentes valeurs des valeurs absolues
solution: x=6 ou x =2/3
3.6)
1er cas
x²-4x-5≥1
x²-4x-6≥1
utiliser méthode delta pour factoriser
x≤2-√10 ou x≥2+√10
2nd cas
-(x²- 4x-5) ≥1
-x²+ 4x +4≥0
2 - 2√2 ≤ x ≤ 2 + 2√2
en définitive
solution :
2 - 2
√2 ≤ x ≤ 2 + 2√2
2 votes
Thanks 1
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Bonsoir j'ai un exercice de math sur les vecteur et je n'arrive pas c'est pour deman c'est urgent merci de votre aide en avance. l'exercice est sur le document ci-joint.
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Lista de comentários
3.1)
|x+7|≥3
1er cas
x+7 ≥ 0 => x+7 ≥3 => x ≥ 3-7 => x ≥ - 4
2nd cas
x+7 ≤ 0 => - (x +7 ) ≥ 3 => -x -7≥3 => x ≤ -7-3 => x ≤ -10
solution :
x≤ -10 OU x ≥ -4
3.2)
|3x-1| ≤5/2
1er cas
3x-1≥0 => 3x-1 ≤ 5/2 => 3x ≤ 5/2 +2/2 => 3x ≤ 7/2
x ≤ 7/6
2nd cas
3x-1≤0 => -(3x-1) ≤ 5/2 => -3x +1 ≤ 5/2
-3x ≤ 5/2 -2/2
-3x ≤ 3/2
3x ≥ -3/2
x ≥ - 3/6 => x ≥ -1/2 ( on simplifie 3/6 = 1/2)
solution
-1/2 ≤ x ≤ 7/6
3.3)
|x+6| ≤0
1er cas
x+6 ≤0 => x ≤ -6
2nd cas
-x-6 ≤0 => x ≥-6
1 seule solution possible x= - 6
3. 4)
|x| +x = 0
1er cas
x>0
x+x = 0 => 2x =0 => x =0
2nd cas
si x <0
-x + x = 0 0 = 0 toujours vrai
donc solution :
x ≤ 0
3.5)
il faut faire un tableau avec les différentes valeurs des valeurs absolues
solution: x=6 ou x =2/3
3.6)
1er cas
x²-4x-5≥1
x²-4x-6≥1
utiliser méthode delta pour factoriser
x≤2-√10 ou x≥2+√10
2nd cas
-(x²- 4x-5) ≥1
-x²+ 4x +4≥0
2 - 2√2 ≤ x ≤ 2 + 2√2
en définitive
solution :
2 - 2√2 ≤ x ≤ 2 + 2√2