Répondre :
Bonjour,
On utilise l'identité remarquable a²-b² = (a+b)(a-b) :
[tex]\left(x^2+3x-1\right)^2 - \left(x^2-3x+1\right)^2 = 0\\ \left[\left(x^2+3x-1\right)-\left(x^2-3x-1\right)\right]\left[\left(x^2+3x-1\right)+\left(x^2-3x-1\right)\right]\\ \left(x^2+3x-1-x^2+3x+1\right)\left(x^2+3x-1+x^2-3x-1\right) = 0\\ 6x\left(2x^2-2\right) = 0\\[/tex]
On continue à factoriser :
[tex]6x\left(2x^2-2\right) = 0\\ 12x\left(x^2-1\right) = 0\\ 12x\left(x+1\right)\left(x-1\right) = 0[/tex]
Donc :
12x = 0
x = 0
OU
x+1 = 0
x = -1
OU
x-1 = 0
x = 1
[tex]S = \left\{-1 ; 0 ; 1\right\}[/tex]
On utilise l'identité remarquable a²-b² = (a+b)(a-b) :
[tex]\left(x^2+3x-1\right)^2 - \left(x^2-3x+1\right)^2 = 0\\ \left[\left(x^2+3x-1\right)-\left(x^2-3x-1\right)\right]\left[\left(x^2+3x-1\right)+\left(x^2-3x-1\right)\right]\\ \left(x^2+3x-1-x^2+3x+1\right)\left(x^2+3x-1+x^2-3x-1\right) = 0\\ 6x\left(2x^2-2\right) = 0\\[/tex]
On continue à factoriser :
[tex]6x\left(2x^2-2\right) = 0\\ 12x\left(x^2-1\right) = 0\\ 12x\left(x+1\right)\left(x-1\right) = 0[/tex]
Donc :
12x = 0
x = 0
OU
x+1 = 0
x = -1
OU
x-1 = 0
x = 1
[tex]S = \left\{-1 ; 0 ; 1\right\}[/tex]