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Titanfal77oyu2hi
@Titanfal77oyu2hi
May 2019
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Voila, qui peut m'aider svp, c'est tres important merci (Math 3eme)
Exercice en piece jointe
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Commentaires (1)
Bonjour
f(x) = 2 x - 7
f(1) = 2 - 7 = - 5
g(x) = x²
g(1) = 1
h(x) = 1 / ( x - 3)
h( 1) = 1 / ( 1 - 3) = 1 /-2
f(0) = - 7
g (0) = 0
h(0) = - 1/3
tu peux maintenant ranger
f(2) = 4 - 7 = - 3
g(2) = 4
h(x) = 1/- 1 = - 1
tu ranges
g(-1) = 1
h(-1) = 1/ ( -1 -3) = 1/-4 = - 1/4
f(x) = 1
2 x - 7 = 1
2 x = 1 +7
2 x = 8
x = 4
h(x) = 1
1 ( x -3) = 1
1 ( x -3) = ( x -3) / x -3)
1 = x - 3
1 + 3 = x
x = 4
3 ne possède pas d'antécédent par h car le dénominateur serait nul , or il est impossible de diviser par 0
1 votes
Thanks 1
titanfal77oyu2hi
Merci a toi, tous n'est pas complet mais je comprend mieux l'exo. un grand merci
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Lista de comentários
f(x) = 2 x - 7
f(1) = 2 - 7 = - 5
g(x) = x²
g(1) = 1
h(x) = 1 / ( x - 3)
h( 1) = 1 / ( 1 - 3) = 1 /-2
f(0) = - 7
g (0) = 0
h(0) = - 1/3
tu peux maintenant ranger
f(2) = 4 - 7 = - 3
g(2) = 4
h(x) = 1/- 1 = - 1
tu ranges
g(-1) = 1
h(-1) = 1/ ( -1 -3) = 1/-4 = - 1/4
f(x) = 1
2 x - 7 = 1
2 x = 1 +7
2 x = 8
x = 4
h(x) = 1
1 ( x -3) = 1
1 ( x -3) = ( x -3) / x -3)
1 = x - 3
1 + 3 = x
x = 4
3 ne possède pas d'antécédent par h car le dénominateur serait nul , or il est impossible de diviser par 0