Bonjour
1) Développer A
A(x) = 2(x - 1)² - (x - 1)(x + 3) + 3
A(x) = 2(x² -2x + 1) - (x² +3x - x - 3)+3
A(x) = 2x² -4x + 2 - x² -3x + x + 3 + 3
A(x) = x² -4x + 2 -2x + 6
A(x) = x² -6x + 8
2)
a)
A(x) = (x - 3)² - 1
A(x) = (x² - 6x + 9) - 1
A(x) = x² - 6x + 8
on retrouve bien l'expression A(x) du 1)
b)
A(x) = (x - 3)² - 1
(x - 3)² = 1
(x - 3) = + √1
x = 1 + 3
x = 4
ou
(x - 3) = - √1
(x - 3) = - 1
x = - 1 + 3
x = 2
donc A(x) = (x - 2)(x - 4)
3)
a)
A(3) = (3 - 2)(3 - 4)
A(3) = (1)(-1)
A(3) = -1
b)
A(-1-√7) = (-1-√7 - 2)(-1-√7 - 4)
A(-1-√7) = (-3 - √7)(-5 - √7)
A(-1-√7) = +15 + 3√7 + 5√7 + 7
A(-1-√7) = +22 + 3√7 + 5√7
A(-1-√7) = +22 + √7 ( 3 + 5)
A(-1-√7) = +22 + 8√7
4)
a)
A(x) = 8
(x - 3)² - 1 = 8
(x - 3)² - 1 - 8 =0
(x - 3)² - 1 - 8 =0
(x - 3)² - 9 =0 Forme A² - B²
( (x - 3) - 3)( (x - 3) + 3) =0
( x - 6)( x )=0
deux solutions S={0, 6}
b)
Résoudre A(x) = 1
(x - 3)² - 1 = 1
(x - 3)² - 2 = 0
( (x - 3) - √2)( (x - 3) + √2) =0
( x - 3 - √2)( x - 3 + √2) =0
Deux solutions :
( x - 3 - √2) = 0
x = + 3 + √2
ou
( x - 3 + √2) =0
x = 3 - √2
S={3 - √2, + 3 + √2}
c)
Résoudre A(x) = -6x
x² - 6x + 8 = - 6x
x² - 6x +6x + 8 = 0
x² + 8 = 0
x² = -8
impossible une racine carré est toujours positive.
donc pas de solution S= { ∅ }
Bon courage
