LALOUCHEA
Answered

Bonjour Je dois résoudre les d'équations en les ramenant à une équation produit nul comment faire?

merci de votre réponse.


[tex]a. \\ x(2x - 5) = 2(2x - 5)[/tex]
[tex]b. \\ (5x + 2)(2x - 3) = 4x {}^{2} - 12x + 9[/tex]
[tex]c. \\ 5x {}^{2} = 8x(x + 1)[/tex]
[tex]d. \\ (2x + 1) {}^{2} = 9[/tex]
[tex]e. \\ (4x + 7) {}^{2} = (x - 2) {}^{2} [/tex]
[tex]f. \\ (7x - 2) {}^{2} - 1 = 80[/tex]

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Sagot :

LOULAKAR

Réponse :

Explications étape par étape

Bonsoir

x(2x - 5) = 2(2x - 5)

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

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

2x - 5 = 0 ou x - 2 = 0

2x = 5 ou x = 2

x = 5/2 ou x = 2

x = 2,5 ou x = 2

(5x + 2)(2x - 3) = 4x^2 - 12x + 9

(5x + 2)(2x - 3) - [(2x)^2 - 2 * 2x * 3 + 3^2] = 0

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

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

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

2x - 3 = 0 ou 3x + 5 = 0

2x = 3 ou 3x = -5

x = 3/2 ou x = -5/3

x = 1,5 ou x = -5/3

5x^2 = 8x(x + 1)

5x^2 - 8x(x + 1) = 0

x[5x - 8(x + 1)] = 0

x(5x - 8x - 8) = 0

x(-3x - 8) = 0

x = 0 ou -3x - 8 = 0

x = 0 ou 3x = -8

x = 0 ou x = -8/3

(2x + 1)^2 = 9

(2x + 1)^2 - 9 = 0

(2x + 1)^2 - 3^2 = 0

(2x + 1 - 3)(2x + 1 + 3) = 0

(2x - 2)(2x + 4) = 0

2(x - 1) * 2(x + 2) = 0

4(x - 1)(x + 2) = 0

x - 1 = 0 ou x + 2 = 0

x = 1 ou x = -2

(4x + 7)^2 = (x - 2)^2

(4x + 7)^2 - (x - 2)^2 = 0

(4x + 7 - x + 2)(4x + 7 + x - 2) = 0

(3x + 9)(5x + 5) = 0

3(x + 3) * 5(x + 1) = 0

15(x + 3)(x + 1) = 0

x + 3 = 0 ou x + 1 = 0

x = -3 ou x = -1

(7x - 2)^2 - 1 = 80

(7x - 2)^2 - 1 - 80 = 0

(7x - 2)^2 - 81 = 0

(7x - 2)^2 - 9^2 = 0

(7x - 2 - 9)(7x - 2 + 9) = 0

(7x - 11)(7x + 7) = 0

(7x - 11) * 7(x + 1) = 0

7(7x - 11)(x + 1) = 0

7x - 11 = 0 ou x + 1 = 0

7x = 11 ou x = -1

x = 11/7 ou x = -1