Réponse:
oui
Explications étape par étape:
1 . C. f (x) = 1 = 3x²+5x+3
1-3x²+-x-3 = 0
-2-3x²-5x = 0
2+3x²+5x = 0
3x²+2x+3x+2 = 0
3x(x+1)+2(x+1) = 0
(x+1)(3x+2) = 0
x1 = x+1
x1 : -1
et
x2 = 3x+2
x2 : -(2/3)
f (x) = 3 = 3x²+5x+3
0 = 3x²+5x
0-3x²-5x = 0
-x(3x+5) = 0
x(3x+5) = 0
x1 : x = 0
x1 = 0
et
x2 : 3x+5 = 0
x2 = -(5/3)
2. a. f(x) = (2x+5)²-1
4x²+20x+25-1
4x²+20x+24
b. f(x) = (2x+5)²-1
(2x+5-1)(2x+5+1)
(2x+4)(2x+6)
2(x+2)×2(x+3)
2×2(x+2)(x+3)
4(x+2)(x+3)
c. f(x) = 0 = 4x²+20x+24
4x²+20x+24 = 0
x²+5x+6 = 0
x²+3x+2x+6 = 0
x(x+3)+2(x+3) = 0
(x+3)(x+2) = 0
x1 : x+3 = 0
x1 = -3
et
x2 : x+2 = 0
x2 = -2
f(x) = 24 = 4x²+20x+24
0 = 4x²+20x
-4x(x+5) = 0
x(x+5) = 0
x1 : x = 0
x1 = 0
et
x2 : x+5 = 0
x2 = -5
f(x) = 3 = 4x²+20x+24
3-4x²+20x+24 = 0
-21-4x²-20x = 0
4x²+20x+21 = 0
4x²+14x+6x+21 = 0
2x(2x+7)+3(2x+7) = 0
(2x+7)(2x+3) = 0
x1 : 2x+7 = 0
x1 = -(7/2)
et
x2 : 2x+3 = 0
x2 = -(3/2)
3. a. x²-6x+5 = (x-1)(x-5)
= x²-5x-1x+5
= x²-6x+5
b. f(x) > 3
x²-6x+8 > 3
x²-6x+5 > 0
x(x-1)-5(x-1) > 0
(x-1)(x-5) > 0
on résout les inégalités
x-1 > 0 = x > 1
x+5 > 0 = x > 5
et
x-1 < 0 = x < 1
x+5 < 0 = x < 5
Soit
x E ] 5;+infini [
x E ] -infini;1 [
Donc
x E ] -infini;1[ U ] 5;+infini [
