Réponse :
Exercice 4.
Résoudre les équations suivantes :
1. x-5√x+4=0 il faut que x ≥ 0
⇔ x + 4 = 5√x
⇔ (x + 4)² = (5√x)²
⇔ x² + 8x + 16 = 25x
⇔ x² - 17x + 16 = 0
Δ = 289 - 64 = 225 > 0
x1 = 17+15)/2 = 16
x2 = 17 - 15)/2 = 1
donc S = {1 ; 16}
2. -x⁴ +3x²-2=0
on pose X = x²
- X² + 3X - 2 = 0
Δ = 9 - 8 = 1 > 0
X1 = - 3 + 1)/-2 = 1 ⇔ x² = 1 ⇔ x = - 1 ou x = 1
X2 = - 3 - 1)/-2 = 2 ⇔ x² = 2 ⇔ x = - √2 ou x = √2
donc S = {- √2 ; - 1 ; 1 ; √2}
3. 6/x² + 1/x -2=0 il faut que x ≠ 0
⇔ 6/x² + x/x² - 2x²/x² = 0
⇔ 6 + x - 2x² = 0
⇔ - 2x² + x + 6 = 0
Δ = 1 + 48 = 49 > 0
x1 = - 1 + 7)/-4 = - 3/2
x2 = - 1 - 7)/-4 = 2
4. 2(cos x)² +3 cosx - 2=0
on pose X = cosx
2X² + 3X - 2 = 0
Δ = 9 + 16 = 25 > 0
X1 = - 3 + 5)/4 = 1/2 ⇔ cosx = 1/2 ⇔ S = {π/3 + 2kπ ; 5π/3 + 2kπ k∈Z}
X2 = - 3 - 5)/4 = - 2 ⇔ cosx = - 2 pas de solutions car - 1 ≤ cosx ≤ 1
5. (x²-3x+1)²-3(x²-3x+1)+2=0
Explications étape par étape :
