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Dorcase789
@Dorcase789
May 2019
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Bonjour pouvez vous m'aider à faire cet exercice dont voici l'énoncé
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scoladan
Verified answer
Bonjour,
1)
2x² + x - 1 = (x + 1).P
⇒ P = 2x - 1
(x + 1)(2x - 1) = 2x² - x + 2x - 1 = 2x² + x - 1
2)
(2x² + x - 1)/(x² - 4)
= (2x - 1)(x + 1)/(x - 2)(x + 2)
x -∞ -2 -1 1/2 2 +∞
2x-1 - - - 0 + +
x+1 - - 0 + + +
x-2 - - - - 0 +
x+2 - 0 + + + +
+ 0 - 0 + 0 - 0 +
3)
(5x² + x - 13)/(x² - 4) ≤ 3
⇔ (5x² + x - 13)/(x² - 4) - 3 ≤ 0
⇔ [(5x² + x - 13) - 3(x² - 4)]/(x² - 4) ≤ 0
⇔ (2x² + x - 1)/(x² - 4) ≤ 0
D'après le tableau de signes du 2) :
S = [-2;-1]U[1/2;2]
2 votes
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Verified answer
Bonjour,1)
2x² + x - 1 = (x + 1).P
⇒ P = 2x - 1
(x + 1)(2x - 1) = 2x² - x + 2x - 1 = 2x² + x - 1
2)
(2x² + x - 1)/(x² - 4)
= (2x - 1)(x + 1)/(x - 2)(x + 2)
x -∞ -2 -1 1/2 2 +∞
2x-1 - - - 0 + +
x+1 - - 0 + + +
x-2 - - - - 0 +
x+2 - 0 + + + +
+ 0 - 0 + 0 - 0 +
3)
(5x² + x - 13)/(x² - 4) ≤ 3
⇔ (5x² + x - 13)/(x² - 4) - 3 ≤ 0
⇔ [(5x² + x - 13) - 3(x² - 4)]/(x² - 4) ≤ 0
⇔ (2x² + x - 1)/(x² - 4) ≤ 0
D'après le tableau de signes du 2) :
S = [-2;-1]U[1/2;2]