Répondre :

[tex](x^2+x-2)(x^2-2x-1)=0\\\\ x^2+x-2 = 0\\ \Delta = 1^2 - 4\times 1\times(-2)\\ \Delta = 9\\\\ x = \frac{-1- \sqrt{9} }{2\times1} = -2 \ \ \ \ \ \ \ \ \ ou \ \ \ \ \ \ \ x = \frac{-1+ \sqrt{9} }{2\times1} =1\\\\ou\\\\ x^2-2x-1=0\\ \Delta = (-2)^2-4\times1\times(-1)\\ \Delta = 8\\\\ x = \frac{2- \sqrt{8} }{2} = 1- \sqrt{2} \ \ \ \ \ ou \ \ \ \ x = \frac{2+ \sqrt{8} }{2} = 1+ \sqrt{2} \\\\ \boxed{S = -2 ; {1- \sqrt{2} ; 1;1+ \sqrt{2} }}[/tex]

D'autres questions