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Louis2006
@Louis2006
April 2019
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Bonjour si quelqu'un pourrait me faire cette exercice cela serais génial merci d'avance
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anylor
* ça veut dire multiplier
on choisit 2
2 + 6
(2 + 6)*2
[(2 + 6)*2] +9
résultat = 8*2 +9 = 25
= 5²
on choisit 5
5 +6
(5+6)*5
[(5+6)*5] +9
résultat = 11 *5 +9 = 55+9 =64
= 8²
c) on choisit x
x +6
(x+6 ) * x
[(x+6 ) * x] +9
résultat = x² + 6x +9
identité remarquable ( x+3)²
image de 2 = f(2) = 25
image de 5 = f(5) = 64
d)
x =2 f(x) = 25 =5²
x =10 f(x) = 169 = 13²
x =0 f(x) = 9 = 3²
x =-15 f(x) = 144 = 12²
x =-8 f(x) = 25 = 5²
x =2.5 f(x) = 30.25 = ( 5.5)²
antécédent de 1 ce sont x = -4 et x = -2
il y en a deux f( -4) = 1 et f(-2) = 1
g) -> tableur
h) on peut choisir 6 ou -12
i) ( x+3)² = 81 = 9²
donc x+3 = 9 => x = 6
2 votes
Thanks 1
louis2006
super merci
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Bonjour aidez-moi (exercice ci-joint) merci d'avance
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Lista de comentários
on choisit 2
2 + 6
(2 + 6)*2
[(2 + 6)*2] +9
résultat = 8*2 +9 = 25
= 5²
on choisit 5
5 +6
(5+6)*5
[(5+6)*5] +9
résultat = 11 *5 +9 = 55+9 =64
= 8²
c) on choisit x
x +6
(x+6 ) * x
[(x+6 ) * x] +9
résultat = x² + 6x +9
identité remarquable ( x+3)²
image de 2 = f(2) = 25
image de 5 = f(5) = 64
d)
x =2 f(x) = 25 =5²
x =10 f(x) = 169 = 13²
x =0 f(x) = 9 = 3²
x =-15 f(x) = 144 = 12²
x =-8 f(x) = 25 = 5²
x =2.5 f(x) = 30.25 = ( 5.5)²
antécédent de 1 ce sont x = -4 et x = -2
il y en a deux f( -4) = 1 et f(-2) = 1
g) -> tableur
h) on peut choisir 6 ou -12
i) ( x+3)² = 81 = 9²
donc x+3 = 9 => x = 6