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Abipipoune06
@Abipipoune06
May 2019
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Bonjour j'ai vraiment besoin d'aide pour mon dm de maths , je suis en terminale L je bloque dessus depuis 2 semaines je n ai pas put assister au cours donc malheureusement je ne comprend pas. S'il vous plaît.
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scoladan
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Bonjour,
1er exo
1)a) 2 - e⁻ˣ⁺¹ = 0
⇔ e⁻ˣ⁺¹ = 2
⇒ -x + 1 = ln(2)
⇔ x = 1 - ln(2)
b) 4ln(x) + 3 = 0
⇔ ln(x) = 3/4
⇒ x = e³/⁴
c) (eˣ - 9)(ln(x) + 1) = 0
⇒ eˣ - 9 = 0 ou ln(x) + 1 = 0
⇔ x = ln(9) = 2ln(3) ou x = e⁻¹ = 1/e
d) e²ˣ - 3eˣ ≥ 0
⇔ eˣ(eˣ - 3) ≥ 0
⇒ eˣ - 3 ≥ 0 car eˣ > 0
⇔ eˣ ≥ 3
⇔ x ≥ ln(3)
2) a)
A = ln[(3 x 5²)/27]
= ln(3) + ln(5²) - ln(27)
= ln(3) + 2ln(5) - 3ln(3) (27 = 3³)
= ln(5) - 2ln(3)
b) B = 3ln(10) + ln(0,08) - 5ln(2)
= 3ln(10) + ln(8 x 10⁻²) - 5ln(2)
= 3ln(10) + ln(8) + ln(10⁻²) - 5ln(2)
= 3ln(10) + 3ln(2) - 2ln(10) - 5ln(2)
= ln(10) - 2ln(2)
= ln(2 x 5) - 2ln(2)
= ln(2) + ln(5) - 2ln(2)
= ln(5) - ln(2)
= ln(5/2)
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Bonjour,1er exo
1)a) 2 - e⁻ˣ⁺¹ = 0
⇔ e⁻ˣ⁺¹ = 2
⇒ -x + 1 = ln(2)
⇔ x = 1 - ln(2)
b) 4ln(x) + 3 = 0
⇔ ln(x) = 3/4
⇒ x = e³/⁴
c) (eˣ - 9)(ln(x) + 1) = 0
⇒ eˣ - 9 = 0 ou ln(x) + 1 = 0
⇔ x = ln(9) = 2ln(3) ou x = e⁻¹ = 1/e
d) e²ˣ - 3eˣ ≥ 0
⇔ eˣ(eˣ - 3) ≥ 0
⇒ eˣ - 3 ≥ 0 car eˣ > 0
⇔ eˣ ≥ 3
⇔ x ≥ ln(3)
2) a)
A = ln[(3 x 5²)/27]
= ln(3) + ln(5²) - ln(27)
= ln(3) + 2ln(5) - 3ln(3) (27 = 3³)
= ln(5) - 2ln(3)
b) B = 3ln(10) + ln(0,08) - 5ln(2)
= 3ln(10) + ln(8 x 10⁻²) - 5ln(2)
= 3ln(10) + ln(8) + ln(10⁻²) - 5ln(2)
= 3ln(10) + 3ln(2) - 2ln(10) - 5ln(2)
= ln(10) - 2ln(2)
= ln(2 x 5) - 2ln(2)
= ln(2) + ln(5) - 2ln(2)
= ln(5) - ln(2)
= ln(5/2)