Bonjour
f(x)= (3x-5) (5x-4) + (5-3x) (x+2)
g(x)= 9x²-30x+25
1) resoudre l'equation: f (x) = g(x)
(3x-5) (5x-4) + (5-3x) (x+2) = 9x²-30x+25
(3x - 5)(5x - 4) - (3x - 5)(x + 2) = (3x)^2 - 2 * 3x * 5 + 5^2
(3x - 5)(5x - 4 - x - 2) = (3x - 5)^2
(3x - 5)(4x - 6) = (3x - 5)^2
(3x - 5) * 2(2x - 3) = (3x - 5)^2
2(3x - 5)(2x - 3) - (3x - 5)^2 = 0
(3x - 5)[2(2x - 3) - 3x + 5] = 0
(3x - 5)(4x - 6 - 3x + 5) = 0
(3x - 5)(x - 1) = 0
3x - 5 = 0 ou x - 1 = 0
3x = 5 ou x = 1
x = 5/3 ou x = 1
2) Soit l'expression fractionnair: h(x) = f(x)÷g(x)
a) pour quelles valeur du réel x, h(x) a t-elle un sens?
9x²-30x+25 # 0
(3x - 5)^2 # 0
3x - 5 # 0
3x # 5
x # 5/3
b) Simplifie h(x)
h(x) = (3x - 5)(4x - 6) / (3x - 5)^2
h(x) = (4x - 6)/(3x - 5)
