Bonjour,
Résoudre:
a) x² ≤ 3x
x² -3x ≤ 0
x(x-3) ≤ 0
x= 0 et x= 3
S= [ 0; 3 ]
b) (3x + 1)x < (3x + 1)(6x + 1)
(3x + 1)x - (3x + 1)(6x + 1) < 0
(3x+1)(x-6x-1) < 0
(3x+1)(-5x-1) < 0
x= -1/3 et -5x-1= 0 => -5x= 1 => x= -1/5
S= ] -∞; -1/3 [ U ] -1/5; + ∞ [
c) (x - 1)(2x+4) ≤ x² - 2
(x - 1)(2x+4) ≤ x² - 2
2x²-2x+4x-4 -x²+2 ≤ 0
x²+2x-2 ≤ 0
Δ= 2²-4(1)(-2)= 4+8= 12
x1= (-2-√12)/2= (-2-√(4x3))/2= (-2-2√3)/2= -1-√3
x2= -1+√3
S= [ -1-√3; -1+√3 ]
