bonjour
f (x) = x ( x + 2 ) - ( 2 x - 1 ) ( x + 2 )
f (x) = x² + 2 x - ( 2 x² + 4 x - x - 2 )
f (x) = x² + 2 x - 2 x² - 4 x + x + 2
f (x ) = - x² - x + 2
g (x) = ( 2 x + 3 )² - ( x + 1 )²
g (x) = 4 x² + 12 x + 9 - ( x² + 2 x + 1 )
g (x) = 4 x² + 12 x + 9 - x² - 2 x - 1
g (x) = 3 x² + 10 x + 8
2. f(x) = ( x + 2 ) ( x - 2 x + 1 )
f (x) = ( x + 2 ) ( - x + 1 )
g (x) = ( 2 x + 3 + x + 1 ) ( 2 x + 3 - x - 1 )
g (x) = ( 3 x + 4 ) ( x + 2 )
3 . f ( √3 ) = - ( √3)² - √3 + 2 = - 3 - √3 + 2 = - 1 - √3
g ( √5) = 3 ( √5)² + 10 √5 + 8 = 15 + 10 √5 + 8 = 23 + 10 √5
4 ) f (x) = 2
- x² - x + 2 = 2
- x² - x + 2 - 2 = 0
- x ( x + 1 ) = 0
x = 0 ou - 1
g (x) = 0
( 3 x + 4 ) ( x + 2 ) = 0
x = - 4/3 ou - 2
f (x) = g (x )
- x² - x + 2 = 3 x² + 10 x + 8
- x² - 3 x² - x - 10 x = 8 - 2
- 4 x² - 11 x = 6
- 4 x² - 11 x - 6 = 0
Δ = ( - 11 )² - 4 ( - 4 * - 6 ) = 121 - 96 = 25 = 5 ²
x 1 = ( 11 - 5 ) / - 8 = - 6 /8 = - 3 /4
x 2 = ( 11 + 5 ) / - 8 = - 16/8 = - 2
- 4 ( x + 3 /4 ) ( x - 2 ) = 0
x = - 3/4 ou 2
g ( x ) < 8
3 x² + 10 x + 8 < 8
3 x² + 10 x + 8 - 8 < 0
3 x² + 10 x < 0
x ( 3 x + 10 ) < 0
s'annule en 0 et - 10 /3
x - ∞ - 10/3 0 + ∞
x - - 0 +
3 x + 10 - 0 + +
total + 0 - 0 +
] - 10 /3 ; 0 [
f (x) ≥ 0
f (x) = ( x + 2 ) ( - x + 1 ) ≥ 0
s'annule en - 2 et 1
x - ∞ - 2 1 + ∞
x + 2 - 0 + +
- x + 1 + + 0 -
produit - 0 + -
[ - 2 ; 1 ]
f (x ) < g (x)
les solutions sont à déduire de f (x) = g (x )
bonne journée
