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lyingspll
@lyingspll
January 2021
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Bonjour j'ai un exercice à faire sur maths je bloque un peu
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scoladan
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Bonjour,
Partie A
1) f(-1) = -(-1)³ - 3(-1)² + 13(-1) + 15 = 1 - 3 - 13 + 15 = 0
2) (x + 1)(ax² + bx + c)
= ax³ + (a + b)x² + (b + c)x + c
Donc par analogie des coefficients des termes de même degré :
a = -1
a + b = -3
b + c = 13
c = 15
⇔ a = -1
b = -2
c = 15
Donc f(x) = (x + 1)(-x² - 2x + 15)
3) a) Δ = 4 - 4(-1)(15) = 64 = 8²
donc 2 racines : x = (2 - 8)/-2 = 3 et x = -5
⇒ f(x) = -(x + 1)(x - 3)(x + 5)
b)
x -∞ -5 -1 3 +∞
x + 5 - 0 + + +
x + 1 - - 0 + +
x - 3 - - - 0 +
f(x) + 0 - 0 + 0 -
f(x) > 0 ⇒ x ∈ ]-∞;-5[∪]-1;3[
Partie B
(E) : x³ + 3x - 4 = 0
1) 1 est racine évidente : 1³ + 3 - 4 = 0
2) (x - 1)(ax² + bx + c)
= ax³ + (b - a)x² + (c - b)x - c
⇒ a = 1
b - a = 0
c - b = 3
-c = -4
⇔ a = 1
b = 1
c = 4
⇒ (E) : (x - 1)(x² + x + 4)
3) Δ = 1² - 4x1x4 < 0 donc pas de racine
Donc solutions de E : x = 1
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Bonjour, voir mon document Merci.
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Bonjour, Est ce que vous pouvez m'aider en anglais, la consigne est dans ci-joint
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lyingspll
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Lista de comentários
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Bonjour,Partie A
1) f(-1) = -(-1)³ - 3(-1)² + 13(-1) + 15 = 1 - 3 - 13 + 15 = 0
2) (x + 1)(ax² + bx + c)
= ax³ + (a + b)x² + (b + c)x + c
Donc par analogie des coefficients des termes de même degré :
a = -1
a + b = -3
b + c = 13
c = 15
⇔ a = -1
b = -2
c = 15
Donc f(x) = (x + 1)(-x² - 2x + 15)
3) a) Δ = 4 - 4(-1)(15) = 64 = 8²
donc 2 racines : x = (2 - 8)/-2 = 3 et x = -5
⇒ f(x) = -(x + 1)(x - 3)(x + 5)
b)
x -∞ -5 -1 3 +∞
x + 5 - 0 + + +
x + 1 - - 0 + +
x - 3 - - - 0 +
f(x) + 0 - 0 + 0 -
f(x) > 0 ⇒ x ∈ ]-∞;-5[∪]-1;3[
Partie B
(E) : x³ + 3x - 4 = 0
1) 1 est racine évidente : 1³ + 3 - 4 = 0
2) (x - 1)(ax² + bx + c)
= ax³ + (b - a)x² + (c - b)x - c
⇒ a = 1
b - a = 0
c - b = 3
-c = -4
⇔ a = 1
b = 1
c = 4
⇒ (E) : (x - 1)(x² + x + 4)
3) Δ = 1² - 4x1x4 < 0 donc pas de racine
Donc solutions de E : x = 1