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NicolasPERE
@NicolasPERE
January 2021
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c'est le 109 svp urgent
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nadiatukiqi
1. (2x+1)^2 - (x-3) (2x+1) = (2x)^2 + 2*2x*1 + 1^2 - (2x^2+x- 6x-3) = 4x^2 + 4x + 1 - 2x^2 - x + 6x -3 = 2x^2 + 9x + 4 2. 2x^2 + 9x + 4 = (2x+1) [ ( 2x + 1) - (x-3) ] = (2x+1) (x+4) 3. On veut résoudre l'équation D= 0 (2x+1) (x+4) = 0 C'est une équation produit nul Si A*B = 0, alors A= 0 ou B = 0 • 2x + 1 = 0 2x = -1 x = -1/2 x = -0,5 • x+4 = 0 x = -4 Cette équation à deux solutions : -0,5 et -4 . 4. On veut résoudre l'équation D = 4 2x^2 + 9x + 4 = 4 On enlève 4 2x^2 + 9x = 0 On factorise 2x^2 + 9x : x(2x+9) x(2x+9) = 0 C'est une équation produit nul . • x = 0 • 2x+9 = 0 2x = -9 x = -9/2 x = -4,5 Il y a deux solutions : 0 et -4,5 5. Lorsque x = -3/2 2x^2 + 9x + 4 = 2 * (-3/2)^2 + 9*-3/2 + 4 = 2* (-9/4) -27/2 +4 = -9/2 - 27/2 + 8/2 = -36/2 + 8/2 = -18 + 4 = -14
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NicolasPERE
ok merci mais j'ai du mal a voir les etape comme 1. 2. 3. 4. 5.
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