Répondre :
a) (2x-1)²-3(2x-1)(x+2) = 7
: 2x²+2*2*1+1²+(-6x+3)(x+2) = 7
: 2x²+5+(-6x²-12x+3x+6) = 7
: 2x²+5-6x²-9x+6 = 7
: -4x²-9x+11 = 7
: -4x²-9x+4 = 0
Recherche de Δ : b²-4ac avec a=(-4), b=(-9) et c=4
: (-9)²-4*(-4)*4 = 145
Or, Δ>0 donc l'équation admet deux racines distinctes
x1 : (-b-√Δ)/2a x2 : (-b+√Δ)/2a
: 9-√145/2*(-4) : 9+√145/2*(-4)
x1 = 9-√145/(-8) et x2 = 9+√145/(-8)
b) (x-2)(2x+7) = x²-4
: 2x²+7x-4x-14 = x²-4
: 2x²+3x-14 = x²-4
: x²+3x-10 = 0
Δ = 3²-4*1*(-10) = 49
x1 : -3-7/2 x2 : -3+7/2
x1 = (-5) et x2 = 2
c) (4x-3)/3 = 1-(5x-12)/6
: (4x-3)/3 = -5x+12+6/6
: (4x-3)/3 = (-5x+18)/6
: (4x-3)/3-(-5x+18)/6 = 0
: (8x-6)/6+(5x-18)/6 = 0
: 13x-24/6 = 0
: 13x = 24
: x = (24/13)
d) 1/x(x+1)+1/x(x+3) = 0
: 1/(x²+x)+1/(x²+3x) = 0
: (x²+3x)/(x²+x)(x²+3x)+(x²+x)/(x²+x)(x²+3x) = 0
: (x²+3x)/(x^4+3x³+x³+3x²)+(x²+x)/(x^4+3x³+x³+3x²) = 0
: (x²+3x)/(x^4+4x³+3x²)+(x²+x)/(x^4+4x³+3x²) = 0
: x(x+3)/x(x³+4x²+3x)+x(x+1)/x(x³+4x²+3x = 0
: (x+3)/(x³+4x²+3x)+(x+1)/(x³+4x²+3x) = 0
: 2x+4/(x³+4x²+3x) = 0
: 2x+4 = 0
: 2x = -4
: x = -2
e) (x-1)/(x-3) ≤ (x-2)/(x-4)
: (x-1)/(x-3)-((x-2)/(x-4)) ≤ 0
: (x-1)(x-4)/(x-3)(x-4)-((x-2)(x-3)/(x-4)(x-3)) ≤ 0
: (x²-4x-x+4)/(x²-4x-3x+12)-((x²-3x-2x+6)/(x²-3x-4x+12)) ≤ 0
: (x²-5x+4)/(x²-7x+12)-(x²-5x+6)/(x²-7x+12) ≤ 0
: -2/(x²-7x+12) ≤ 0
: 2/(x²-7x+12) ≥ 0
: ⇔ x²-7x+12 ≥ 0
: Δ = (-7)²-4*1*12 = 1
x1 : 7-1/2 x2 : 7+1/2
x1 = 3 et x2 = 4