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Ornela1803
@Ornela1803
January 2021
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Bonjour ! J'aurais besoin d'aide pour 2 exercices de mathématiques svp . Je n'y comprends rien NIVEAU 1ere S merci d'avance.
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scoladan
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Bonjour,
I)
1) (E) : 10x⁴ - 27x³ - 110x² - 27x + 10 = 0
Pour x = 0 : 10 ≠ 0 donc 0 n'est pas solution
On peut donc diviser par x² :
(E) : 10x² - 27x - 110 - 27/x + 10/x² = 0
2) a) u = x + 1/x
u² - 2 = (x + 1/x)² - 2
= x² + 2 + 1/x² - 2
= x² + 1/x²
b) on en déduit : 10x² + 10/x² = 10(x² + 1/x²) = 10(u² - 2)
et : 27x + 27/x = 27(x + 1/x) = 27u
Soit (E) ⇔ 10(u² - 2) - 27u - 110 = 0
⇔ 10u² - 27u - 130 = 0
c) Δ = (-27)² - 4x10x(-130) = 729 + 5200 = 5929 = 77²
Donc 2 solutions : u₁ = (27 - 77)/20 = -5/2
et u₂ = (27 + 77/20 = 104/20 = 26/5
soit : x + 1/x = -5/2 ou x + 1/x = 26/5
⇒ x² + 5x/2 + 1 = 0 ou x² - 26x/5 + 1 = 0
Δ₁ = (5/2)² - 4 = (3/2)² Δ₂ = (-26/5)² - 4 = 576/25 = (24/5)²
x₁ = (-5/2 - 3/2)/2 = -2 x₃ = (26/5 - 24/5)/2 = 1/5
x₂ = (-5/2 + 3/2)/2 = -1/2 x₄ = (26/5 + 24/5)/2 = 5
II) 1)
A = n² - (n + 1)² - (n + 2)² + (n + 3)²
= n² - (n² + 2n + 1) - (n² + 4n + 4) + (n² + 6n + 9)
= 4
2) B = (1² - 2² - 3² + 4²) + (5² - 6² - 7² + 8²) + .... + (2009² - 2010² - 2011² + 2012²)
= [1² - (1 + 1)² - (1 + 2)² + (1 + 3)²] + ..... + [2009² - (2009 + 1)² - (2009 + 2)² + (2009 + 3)²]
= 4 + 4 + .... + 4 (2012/4 = 503 donc 503 fois)
= 503 x 4
= 2012
2 votes
Thanks 1
Ornela1803
Merci beaucoup
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Verified answer
Bonjour,I)
1) (E) : 10x⁴ - 27x³ - 110x² - 27x + 10 = 0
Pour x = 0 : 10 ≠ 0 donc 0 n'est pas solution
On peut donc diviser par x² :
(E) : 10x² - 27x - 110 - 27/x + 10/x² = 0
2) a) u = x + 1/x
u² - 2 = (x + 1/x)² - 2
= x² + 2 + 1/x² - 2
= x² + 1/x²
b) on en déduit : 10x² + 10/x² = 10(x² + 1/x²) = 10(u² - 2)
et : 27x + 27/x = 27(x + 1/x) = 27u
Soit (E) ⇔ 10(u² - 2) - 27u - 110 = 0
⇔ 10u² - 27u - 130 = 0
c) Δ = (-27)² - 4x10x(-130) = 729 + 5200 = 5929 = 77²
Donc 2 solutions : u₁ = (27 - 77)/20 = -5/2
et u₂ = (27 + 77/20 = 104/20 = 26/5
soit : x + 1/x = -5/2 ou x + 1/x = 26/5
⇒ x² + 5x/2 + 1 = 0 ou x² - 26x/5 + 1 = 0
Δ₁ = (5/2)² - 4 = (3/2)² Δ₂ = (-26/5)² - 4 = 576/25 = (24/5)²
x₁ = (-5/2 - 3/2)/2 = -2 x₃ = (26/5 - 24/5)/2 = 1/5
x₂ = (-5/2 + 3/2)/2 = -1/2 x₄ = (26/5 + 24/5)/2 = 5
II) 1)
A = n² - (n + 1)² - (n + 2)² + (n + 3)²
= n² - (n² + 2n + 1) - (n² + 4n + 4) + (n² + 6n + 9)
= 4
2) B = (1² - 2² - 3² + 4²) + (5² - 6² - 7² + 8²) + .... + (2009² - 2010² - 2011² + 2012²)
= [1² - (1 + 1)² - (1 + 2)² + (1 + 3)²] + ..... + [2009² - (2009 + 1)² - (2009 + 2)² + (2009 + 3)²]
= 4 + 4 + .... + 4 (2012/4 = 503 donc 503 fois)
= 503 x 4
= 2012