Bonsoir.
a.
(x - 1)² = 4
(x - 1)² - 4 = 0
(x - 1)² - 2² = 0
(x - 1 + 2)(x - 1 - 2) = 0
(x + 1)(x - 3) = 0
2 solutions :
x + 1 = 0 <=> x = - 1
x - 3 = 0 <=> x = 3
S = {- 1 ; 3}
b.
(2x - 1)² = 2
(2x - 1)² - 2 = 0
(2x - 1)² - (√2)² = 0
(2x - 1 + √2)(2x - 1 - √2) = 0
2 solutions :
2x - 1 + √2 = 0 <=> 2x = 1 - √2 <=> x = (1 - √2) / 2
2x - 1 - √2 = 0 <=> 2x = 1 + √2 <=> x = (1 + √2) / 2
S = {(1 - √2) / 2 ; (1 + √2) / 2}
c.
(x + 4)² = - 2
Un carre ne pouvant etre negatif, cette equation ne comporte aucune solution.
S = {Ø}
d.
(x + 4)² = (7 - 2x)²
(x + 4)² - (7 - 2x)² = 0 [id. rem. a² - b²]
(x + 4 + 7 - 2x)(x + 4 - 7 + 2x) = 0
(11 - x)(3x - 3) = 0
3(11 - x)(x - 1) = 0
(11 - x)(x - 1) = 0
2 solutions :
11 - x = 0 <=> x = 11
x - 1 = 0 <=> x = 1
S = {1 ; 11}
Bonne soiree !