Bonjour à tous Ex. 14 svp merci d'avance

a) x² - 4x = 0

⇔ x(x - 4) = 0 ← Résultat nul si x = 0 ou (x-4) = 0

x - 4 = 0

x = 4

S = {0;4}

b) 2x² + 3 = 0

⇔ 2(x² + 1.5) = 0 ← Résultat nul si (x² + 1.5) = 0

x² + 1.5 = 0

x² = -1.5

S = {∅}

c) x² - 5 = 0

⇔ (x - √5)(x + √5) = 0 ← Résultat nul si (x - √5) = 0 ou (x + √5) = 0

x - √5 = 0

x = √5

x + √5 = 0

x = -√5

S = {-√5;√5}

d) (2x-5)² - 9 = 0

⇔ ((2x-5) - 3)((2x-5) + 3) = 0 ← Résultat nul si ((2x-5) - 3) = 0 ou ((2x-5) + 3) = 0

((2x-5) - 3) = 0

2x - 5 - 3 = 0

2x - 8 = 0

2x = 8

x = 8/2 = 4

((2x-5) + 3) = 0

2x - 5 + 3 = 0

2x - 2 = 0

2x = 2

x = 2/2 = 1

S = {1;4}

e) -x² + 6x - 9 = 0

⇔ - (x² - 6x + 9) = 0

⇔ - (x-3)² = 0 ← Résultat nul si (x-3) = 0

x - 3 = 0

x = 3

S = {3}

Bonjour,

x²-4x= x(x-1)

x(x-1)=

0x= 0  ou x= 1   S= {0 ; -1}

2x²+3= 0  pas de solutions.   S= { Ф }

(2x-5)²-9= (2x-5)²- 3²

(2x-5)-3²= 0    

(2x-5-3)(2x-5+3)= 0

(2x-8)(2x-2)= 0

x= 4    ou   x= 1    S= {1 ; 4}

-x²+6x-9= -(x-3)²  trouve la ou les solutions .