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TonyStarkDu72
@TonyStarkDu72
May 2019
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Pouver vous resoudre ca svp. cest pas long....
(3-x)^2×2 = (x+9)(x+2)
A savoir que ca ne peut pas etre -9
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Verified answer
2*(3-x)²=(x+9)(x+2)
2*(9-6x+x²)=x²+2x+9x+18
18-12x+2x²=x²+11x+18
x²-23x=0
x*(x-23)=0
soit x=0 ou x-23=0
x=23
0 votes
Thanks 1
TonyStarkDu72
attend je vais vérifier. tes sympa
RomainVTT
Bonsoir,
(3-x)²×2 = (x+9)(x+2)
=> (9 -6x +x²) *2 = x² +2x +9x +18
=> 18 -12x +2x² = x² +11x +18
=> 18 -18 +2x² -x² -12x -11x = 0
=> x² -23x = 0
=> x (x -23) = 0
............pour qu'un produit de facteurs soit nul, il suffit que l'un des facteurs soit nul
donc :
x (x-23) = 0 si x = 0 ou (x -23) = 0 => x = 23
1 votes
Thanks 0
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Verified answer
2*(3-x)²=(x+9)(x+2)2*(9-6x+x²)=x²+2x+9x+18
18-12x+2x²=x²+11x+18
x²-23x=0
x*(x-23)=0
soit x=0 ou x-23=0
x=23
(3-x)²×2 = (x+9)(x+2)
=> (9 -6x +x²) *2 = x² +2x +9x +18
=> 18 -12x +2x² = x² +11x +18
=> 18 -18 +2x² -x² -12x -11x = 0
=> x² -23x = 0
=> x (x -23) = 0
............pour qu'un produit de facteurs soit nul, il suffit que l'un des facteurs soit nul
donc :
x (x-23) = 0 si x = 0 ou (x -23) = 0 => x = 23