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Lamarocainemrk
@Lamarocainemrk
April 2019
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Sans calculer les racine de α et β de l'équation x²+√2x -√6 , calculer :
α+β ; α.β ; α²+β² ; 1/α +1/β ; α³ +β³
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kvnmurty
f(x) = a x² + b x + c = 0 , pour cette equation quadratique, les racines sont:
α = [ - b + √(b² - 4ac) ] / (2 a) et β = [ -b - √(b² - 4ac) ] / (2a)
On peut developper/reduire/deduire les expressions comme suivant:
α + β = - b / a αβ = c/a
α² + β² = (α + β)² - 2 α . β = b²/a² - 2 c/a = (b² - 2 a c) /a²
1/α + 1/β = (α + β) / (α.β ) = - b/c
α³ + β³ = (α + β) ( α² - αβ + β²) = (-b/a) * [ (b² -2ac)/a² - c/a ]
= - b * (b² - 3ac) / a³
==============================
f(x) = x² + √2 x - √6
a = 1 , b = √2 c = -√6
α + β = -√2 / 1 = -√2 α . β = -√6 / 1 = -√6
α² + β² = (2 + 2 √6) /1² = 2 (1+√6)
1/α + 1/β = -√2 /-√6 = 1/√3
α³ + β³ = - √2 [ (2 - 3 * 1* (-√6) ] / 1³
= - √2 (2 + 3√6)
3 votes
Thanks 1
kvnmurty
:)
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SVPPPPP J'AI VRAIMENT BESOIN D'AIDE !!! Montrer que pour tout point M du plan on a : MA.BC + MB.CA + MC.AB = 0(vecteur nul) PS: MA , BC , CA ,MB et AB sont tous des vecteurs !
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lamarocainemrk
February 2021 | 0 Respostas
Montrer que pour tout point M du plan on a : MA.BC + MB.CA + MC.AB = 0(vecteur nul) PS: MA , BC , CA ,MB et AB sont tous des vecteurs !
Responda
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Lista de comentários
α = [ - b + √(b² - 4ac) ] / (2 a) et β = [ -b - √(b² - 4ac) ] / (2a)
On peut developper/reduire/deduire les expressions comme suivant:
α + β = - b / a αβ = c/a
α² + β² = (α + β)² - 2 α . β = b²/a² - 2 c/a = (b² - 2 a c) /a²
1/α + 1/β = (α + β) / (α.β ) = - b/c
α³ + β³ = (α + β) ( α² - αβ + β²) = (-b/a) * [ (b² -2ac)/a² - c/a ]
= - b * (b² - 3ac) / a³
==============================
f(x) = x² + √2 x - √6
a = 1 , b = √2 c = -√6
α + β = -√2 / 1 = -√2 α . β = -√6 / 1 = -√6
α² + β² = (2 + 2 √6) /1² = 2 (1+√6)
1/α + 1/β = -√2 /-√6 = 1/√3
α³ + β³ = - √2 [ (2 - 3 * 1* (-√6) ] / 1³
= - √2 (2 + 3√6)