Réponses

2014-03-17T12:40:22+01:00


1. f(x)= (x-3)(2x+1)-(x-3)²
= (x-3)(2x+1)-(x-3)(x-3)
= (x-3)[(2x+1)-(x-3)]
= (x-3)(2x+1-x+3)
= (x-3)(x+4)

f(x)= (x-3)(2x+1)-(x-3)²
= 2x²+x-6x-3-(x-3)(x-3)
= 2x²-5x-3-x²+6x+9
= x²+x-12

g(x)= (x-1)²-4
= (x-1)²-(2)²
= (x-1-2)(x-1+2)
= (x-3)(x+1)

g(x)= (x-1)²-4
= x²-2x+1-4
= x²-2x-3

2. f(3)=(3-3)(3+4)
= 0*7
= 0
f(√2)= (√2)²+2-12
= 2+√2-12
= √2-10

f(√3+3)= (√3+3)²+√3+3-12
= 3+6√3+9+√3+3-12
= 7√3+3

g(3)= (3-3)(3+1)
= 0*4
= 0

g(√2)= (√2-1)²-4
= 2-2√2+1-4
= -2√2-1

g(√3+3)= (√3+3)²-2(√3+3)-3
= (3+6√3+9)-(2√3+6)-3
= 3+6√3+p-2√3-6-3
= 4√3+3

b. Les antécédents de f par 0 :
f(x) = 0 pour x=3 et x= -4

Les antécédents de g par 0:
g(x)=0 pour x=3 et x=-1