Réponses

2013-09-10T16:17:20+02:00
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+


2013-09-10T16:19:28+02:00
(x+1)^2 - 4
1)  x^2+ 2x+1 - 4
   =x^2 + 2x -3
2) (x+1)^2 - 2^2
  =(x+1+2)(x+1-2)
  =(x+3)(x-1)