Svp aider moi juste la question 4 et 5

 Salut,

1.
f(x) = x(x+2)-(2x-1)(x+2)
f(x) = x² + 2x - (2x² + 4x -x -2) = x² + 2x - 2x² -3x + 2 = -x² -x +2
g(x) = (2x+3)² - (x+1)² = 4x² + 12x + 9 - (x² + 2x+1) = 4x² + 12x + 9 -x² -2x -1 = 3x² +10x+8

2.
f(x) = x(x+2)-(2x-1)(x+2) = (x+2)[x - (2x-1)] = (x+2)(x-2x+1) = (x+2)(-x+1)
g(x) = (2x+3)² - (x+1)²  = [(2x+3)+(x+1)][(2x+3)-(x+1)] = (2x+3+x+2)(2x+3-x-1) = (3x + 5)(x+2)

4.

f(x) = 2
-x² -x +2 = 2
-x² - x + 2 - 2 = 0
-x² -x =0
-x(x+1) = 0
-x = 0
x+1 = 0
x = -1

g(x) = 0
(3x + 5)(x+2) = 0
3x + 5 = 0
3x = -5
x = -5/3

x + 2 = 0
x = - 2


5.

a.

g(x) < 8
3x² +10x+8 < 8
3x² + 10x + 8 - 8 < 0
3x² + 10x < 0
x(3x+10) < 0
x < 0
3x + 10 < 0
3x < -10
x < -10/3

S = ]-infini ; 0[

f(x) ≥ 0

(x+2)(-x+1) ≥ 0
x + 2 ≥0
x ≥ -2
-x + 1 ≥ 0
-x ≥ -1
x ≤ 1

S = [-2;1]

f(x) < g(x)

(x+2)(-x+1) < (3x + 5)(x+2)
(x+2)(-x+1) - (3x+5)(x+2) < 0
(x+2)[(-x+1) - (3x+5)] < 0
(x+2)(-x+1-3x-5) < 0
(x+2)(-4x - 4) < 0
x+ 2 < 0
x < -2
-4x - 4 < 0
-4x < 4
x > -1

S = ]-1; -2[

Bonne soirée !