Réponses

2013-12-13T02:12:08+01:00
Bonsoir,

109 a) f(x) = 2x(x - 7) - x²
f(x) = 2x(x - 7) - x*x
f(x) = x[2(x - 7) - x]
f(x) = x(2x - 14 - x)
f(x) = x(x - 14)

d) f(x) = 3(x + 3)² - 2(x + 3)(x - 5)
f(x) = 3(x + 3)(x + 3) - 2(x + 3)(x - 5)
f(x) = (x + 3)[3(x + 3) - 2(x - 5)]
f(x) = (x + 3)(3x + 9 - 2x + 10)
f(x) = (x + 3)(x + 19)

110) d)  f(x) = (x + 1)(x + 3) - x - 1
f(x) = (x + 1)(x + 3) - (x + 1)
f(x) = (x + 1)(x + 3) - (x + 1)*1
f(x) = (x + 1)[(x + 3) -1]
f(x) = (x + 1)(x + 3 -1)
f(x) = (x + 1)(x + 2)

111) C'est une application de la formule : a² - b² = (a + b)(a - b)

a) f(x) = (x - 3)² - 25
f(x) = (x - 3)² - 5²
f(x) = [(x - 3) +5][(x - 3) - 5]
f(x) = (x - 3 + 5)(x - 3 - 5)
f(x) = (x + 2)(x - 8)

b) f(x) = 4(x - 2)² - 9
f(x) = [2(x - 2)]² - 3²
f(x) = [2(x - 2) + 3][2(x - 2) - 3]
f(x) = (2x - 4 + 3)(2x - 4 - 3)
f(x) = (2x - 1)(2x - 7)

c) f(x) = (x + 1)² - 2
f(x) = (x + 1)^2 - (\sqrt{2})^2\\\\f(x) = [(x + 1) + \sqrt{2}][(x + 1) - \sqrt{2}]\\\\f(x) = (x + 1 + \sqrt{2})(x + 1 - \sqrt{2})

d) f(x) = 5 - (2x - 3)²
f(x)=(\sqrt{5})^2-(2x-3)^2\\\\f(x)=[\sqrt{5}+(2x-3)][\sqrt{5}-(2x-3)]\\\\f(x)=(\sqrt{5}+2x-3)(\sqrt{5}-2x+3)

112) a) f(x) = (x + 5)² - (2x - 7)²
f(x) = [(x + 5) + (2x - 7)][(x + 5) - (2x - 7)]
f(x) = (x + 5 + 2x - 7)(x + 5 - 2x + 7)
f(x) = (3x - 2)(-x + 12)

b) f(x) = 4(x + 1)² - 9x²
f(x) = [2(x + 1)]² - (3x)²
f(x) = [2(x + 1) + 3x][2(x + 1) - 3x]
f(x) = (2x + 2 + 3x)(2x + 2 - 3x)
f(x) = (5x + 2)(-x + 2)

c) f(x) = -4x² + 4x - 1
f(x) = -(4x² - 4x + 1)
f(x) = -(2x - 1)²

d) f(x)=\dfrac{1}{2}x^2+x+\dfrac{1}{2}

f(x)=\dfrac{1}{2}(x^2+2x+1)\\\\f(x)=\dfrac{1}{2}(x+1)^2