Explications étape par étape :
a)
x ( x + 1 ) = 2( x + 1 )
⇔ x ( x + 1 ) - 2( x + 1 ) = 0 factorisation
⇔ ( x + 1 ) ( x - 2 ) = 0 Equation produit nul
Il suffit que :
x + 1 = 0 ou x - 2 = 0
⇔ x = -1 ⇔ x = 2
S = { -1 ; 2 }
b)
( x + 3 ) ( x + 4 ) = ( x + 3 ) ( 2x - 1 )
⇔ ( x + 3 ) ( x + 4 ) - ( x + 3 ) ( 2x - 1 ) = 0
⇔ ( x + 3 ) [ ( x + 4 ) - ( 2x - 1 ) ] = 0
⇔ ( x + 3 ) ( x + 4 - 2x + 1 ) = 0
⇔ ( x + 3 ) ( -x + 5 ) = 0 Equation produit nul
x = -3 ou x = 5
S = { -3 ; 5 }
c)
(x+3) (x - 2) = x² - 4
⇔ (x + 3 ) ( x - 2 ) - ( x² - 4 ) = 0
⇔ (x + 3 ) ( x - 2 ) - ( x - 2 ) ( x + 2 ) = 0
⇔ ( x - 2 ) [ (x + 3 ) - ( x + 2 )] = 0
⇔ ( x - 2 ) ( x + 3 - x - 2 ) = 0
⇔ ( x - 2 ) = 0
S = { 2 }
d)
( x + 3 ) ( x - 5 ) = x² - 9
⇔ ( x + 3 ) ( x - 5 ) - ( x² - 9 ) = 0
⇔ ( x + 3 ) ( x - 5 ) - ( x - 3 ) ( x + 3 ) = 0
⇔ ( x + 3 ) [ ( x - 5 ) - ( x - 3 ) ] = 0
⇔ ( x + 3 ) ( x - 5 - x + 3 ) = 0
⇔ ( x + 3 ) ( -2 ) = 0
S = { -3 }
e)
( x - 8 )² = ( 7 - 2x )²
⇔ ( x - 8 )² - ( 7 - 2x )² = 0 a² - b²
⇔ ( x - 8 - 7 + 2x ) ( x - 8 + 7 - 2x ) = 0
⇔ ( 3x - 15 ) ( -x - 1 ) = 0
S = { -1 ; 5 }
g) i) f) h) j) même procédé que e)
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Explications étape par étape :
a)
x ( x + 1 ) = 2( x + 1 )
⇔ x ( x + 1 ) - 2( x + 1 ) = 0 factorisation
⇔ ( x + 1 ) ( x - 2 ) = 0 Equation produit nul
Il suffit que :
x + 1 = 0 ou x - 2 = 0
⇔ x = -1 ⇔ x = 2
S = { -1 ; 2 }
b)
( x + 3 ) ( x + 4 ) = ( x + 3 ) ( 2x - 1 )
⇔ ( x + 3 ) ( x + 4 ) - ( x + 3 ) ( 2x - 1 ) = 0
⇔ ( x + 3 ) [ ( x + 4 ) - ( 2x - 1 ) ] = 0
⇔ ( x + 3 ) ( x + 4 - 2x + 1 ) = 0
⇔ ( x + 3 ) ( -x + 5 ) = 0 Equation produit nul
Il suffit que :
x = -3 ou x = 5
S = { -3 ; 5 }
c)
(x+3) (x - 2) = x² - 4
⇔ (x + 3 ) ( x - 2 ) - ( x² - 4 ) = 0
⇔ (x + 3 ) ( x - 2 ) - ( x - 2 ) ( x + 2 ) = 0
⇔ ( x - 2 ) [ (x + 3 ) - ( x + 2 )] = 0
⇔ ( x - 2 ) ( x + 3 - x - 2 ) = 0
⇔ ( x - 2 ) = 0
S = { 2 }
d)
( x + 3 ) ( x - 5 ) = x² - 9
⇔ ( x + 3 ) ( x - 5 ) - ( x² - 9 ) = 0
⇔ ( x + 3 ) ( x - 5 ) - ( x - 3 ) ( x + 3 ) = 0
⇔ ( x + 3 ) [ ( x - 5 ) - ( x - 3 ) ] = 0
⇔ ( x + 3 ) ( x - 5 - x + 3 ) = 0
⇔ ( x + 3 ) ( -2 ) = 0
S = { -3 }
e)
( x - 8 )² = ( 7 - 2x )²
⇔ ( x - 8 )² - ( 7 - 2x )² = 0 a² - b²
⇔ ( x - 8 - 7 + 2x ) ( x - 8 + 7 - 2x ) = 0
⇔ ( 3x - 15 ) ( -x - 1 ) = 0
S = { -1 ; 5 }
g) i) f) h) j) même procédé que e)