bonsoir emma
(x-1)
(3x+2) =x²-1
(x-1)
(3x+2) =(x-1)(x+1)
(x-1)
(3x+2) -(x-1)(x+1) =0
(x-1)
[ 3x+ 2 - x -1]
=(x-1)(
2x + 1)
un
produit de facteur est nul, si au moins un de ses facteurs est nul
x-1
=0 => x = 1
ou
2x
+1 = 0 => x = - ½
Solution ={ -1/2 ; 1}
b)
(5x+2)²
-(2x-1)² > 0 on utilise l'identité remarquable a²-b² = (a-b) (a+b)
(5x+2
– 2x +1)( 5x+2 +2x-1) > 0
(3x+3)
(7x+1) > 0
3(x+1)(7x+1)
> 0
tableau
de signes
x+1
> 0 => x > -1
7x+1>0 => x > -1/7
3(x+1)(7x+1)
> 0 =>
solution = ]-∞;-1[ U] -1/7;+∞[
c)
(x-3)
(2x+1) ≤ 3( x-3)
(x-3)
(2x+1) -3( x-3) ≤ 0 on met (x-3) en facteur
(x-3)
(2x+1-3) ≤0
(x-3)
(2x-2) ≤0
2(x-3)(x-1) ≤ 0
x-3≤ 0 => x ≤3
x-1≤0 => x≤ 1
tableau de signes
solution = [
1;3]
