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Puchino
@Puchino
January 2021
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Partie B seulement
Bonjour, j'aurai besoin de votre aide sur toute la partie B. Merci beaucoup d'avance :)
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aur70
Verified answer
Bonjour
f(x)=(2-x)(x+3)+3(2-x)
1/ f(x)=2x+6-x²-3x+6-3x=-x²-4x+12
2/ on factorise
f(x)=(2-x)[(x+3)+3] = (2-x)(x+3+3)=(2-x)(x+6)
3/ 16-(x+2)²=4²-(x+2)²=[4-(x+2)][4+(x+2)]=(4-x-2)(4+x+2)=(2-x)(x+6)=f(x)
4/ a/ f(-2)=16-(-2+2)²=16-0=16
b/ f(1+√2)= -(1+√2)²-4(1+√2)+12 = -(1+2√2+2)-4-4√2+12 = -3-2√2-4-4√2+12 = 5-6√2
c/ f(x)=0 ⇔(2-x)(x+6)=0 ⇔2-x=0 ou x+6=0 ⇔ x=2 ou x=-6
5/ a/
x -∞ -6 2 + ∞
(2-x) + + 0 -
(x+6) - 0 + +
produit - + -
b/ f(x)<0 ⇔(2-x)(x+6)<0 ⇔ x<-6 ou x>2 (lecture dans le tableau de signe)
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Lista de comentários
Verified answer
Bonjourf(x)=(2-x)(x+3)+3(2-x)
1/ f(x)=2x+6-x²-3x+6-3x=-x²-4x+12
2/ on factorise
f(x)=(2-x)[(x+3)+3] = (2-x)(x+3+3)=(2-x)(x+6)
3/ 16-(x+2)²=4²-(x+2)²=[4-(x+2)][4+(x+2)]=(4-x-2)(4+x+2)=(2-x)(x+6)=f(x)
4/ a/ f(-2)=16-(-2+2)²=16-0=16
b/ f(1+√2)= -(1+√2)²-4(1+√2)+12 = -(1+2√2+2)-4-4√2+12 = -3-2√2-4-4√2+12 = 5-6√2
c/ f(x)=0 ⇔(2-x)(x+6)=0 ⇔2-x=0 ou x+6=0 ⇔ x=2 ou x=-6
5/ a/
x -∞ -6 2 + ∞
(2-x) + + 0 -
(x+6) - 0 + +
produit - + -
b/ f(x)<0 ⇔(2-x)(x+6)<0 ⇔ x<-6 ou x>2 (lecture dans le tableau de signe)