Bonsoir.
EX 2
1/ f(x) = (x - 1)² - 4 = x² - 2x + 1 - 4 = x² - 2x - 3.
2/ f(x) = (x - 1)² - 4 = (x - 1)² - 2² = (x - 1 - 2) (x - 1 + 2) = (x - 3) (x + 1).
3/ f(x) = 0 ⇔ (x - 3) (x + 1) = 0
2 solutions :
x - 3 = 0 ⇔ x = 3
x + 1 = 0 ⇔ x = - 1
S = {- 1 ; 3}
f(x) = - 4 ⇔ (x - 1)² - 4 = - 4 ⇔ (x - 1)² = 0
1 solution : x - 1 = 0 ⇔ x = 1
S = {1}
f(x) = - 3 ⇔ x² - 2x - 3 = - 3 ⇔ x² - 2x = 0 ⇔ x (x - 2) = 0
2 solutions :
x = 0
x - 2 = 0 ⇔ x = 2
S = {0 ; 2}
f(x) = - 9 ⇔ x² - 2x - 3 = - 9 ⇔ x² - 2x + 6 = 0
Δ = (- 2)² - 4 * 6 = 4 - 24 = - 20
Δ < 0 donc pas de solutions.
S = {Ф}
f(x) = 165 ⇔ (x - 1)² - 4 = 165 ⇔ (x - 1)² - 169 = 0 ⇔ (x - 1)² - 13² = 0
⇔ (x - 1 - 13) (x - 1 + 13) = 0 ⇔ (x - 14) (x + 12) = 0
2 solutions :
x - 14 = 0 ⇔ x = 14
x + 12 = 0 ⇔ x = - 12
S = {- 12 ; 14}
4/ Tableau de valeurs :
x - 3 - 2 - 1 0 1 2 3 4
f(x) 12 5 0 - 3 - 4 - 3 0 5
EX 2
1/ (7x - 1) / 4 = (2 - 3x) / 5
5 (7x - 1) = 4 (2 - 3x)
35x - 5 = 8 - 12x
35x + 12x = 8 + 5
47x = 13
1 solution :
x = 13/47
S = {13/47}
2/ (5x - 3) (2 - 3x) = 0 (equation-produit nul)
2 solutions :
5x - 3 = 0 ⇔ 5x = 3 ⇔ x = 3/5
2 - 3x = 0 ⇔ 3x = 2 ⇔ x = 2/3
S = {3/5 ; 2/3}
3/ 3 (x - 1)² + 4x - 4 = 0
3 (x - 1)² + 4 (x - 1) = 0
(x - 1) [3 (x - 1) + 4] = 0
(x - 1) (3x - 3 + 4) = 0
(x - 1) (3x + 1) = 0 (equation-produit nul)
2 solutions :
x - 1 = 0 ⇔ x = 1
3x + 1 = 0 ⇔ 3x = - 1 ⇔ x = - 1/3
S = {- 1/3 ; 1}
Bonne nuit !
