Bonjour,
1)
f(x) = (x+1)²-4
f(x) = x²+2x+1-4
f(x) = x²+2x-3
2)
c'est une identité remarquable de la forme a²-b²=(a+b)(a-b)
f(x) = (x+1)²-4 =
f(x) = (x+1+2)(x+1-2) =
f(x) = (x+3)(x-1)
3a)
f(0) --> on prend f(x) = x²+2x-3
f(0) = 0+0-3
f(0) = -3
f(1) --> on prend f(x) = (x+3)(x-1)
f(1) = (1+3)(1-1) = 4*0
f(1) = 0
f(-3) --> on prend f(x) = (x+3)(x-1)
f(-3) = (-3+3)(-3-1) = 0*(-4) =
f(-3) = 0
f(V3) -->on prend f(x) = x²+2x-3
f(V3) = 3+2V3-3 =
f(V3) = 2V3
(V = racine carrée)
3b)
f(x) = (x+3)(x-1)
F(x) = 0 si x+3=0 ou x=-3
F(x) = 0 si x-1=0 ou x=1
S = {-3 ; 1}
J'espère que tu as compris
a+
