Réponses

2012-10-10T18:57:26+02:00

je suppose que c'est (x^3)/-(x^2)=1 => x^3=-x^2 => x^3+x^2=0 => x²(x+1)=0

donc 2 solutions :\left \{ {{x^{2}=0} \atop {x+1=0}} \right.

donc :\left \{ {{x=0} \atop {x=-1}} \right.

S={-1;0}