Mathématiques

Question

On considère l'expression : C=(2x-5)²-(2x-5)(3x+7)
Calculer si x = [tex] \frac{1}{2} [/tex]
Développer et réduire C

2 Réponse

  • C = (2/2-5)²-(2/2-5)(3/2+7)
    C =(1-5)²-(1-5)(8,5)
    C = (-4)²-(-4)(8,5)
    C = 16+34
    C = 50

    C = (2x-5)²-(2x-5)(3x+7)
    C = (4x²-20x+25)-(6x²+14x-15x-35)
    C = 4x²-20x+25-(6x²-x-35)
    C = 4x²-20x+25-6x²+x+35
    C = -2x²-19x+60
  • On considère l'expression : C = (2x - 5)² - (2x - 5) (3x + 7)

    Calculer si x = 1/2
    C = (2x - 5)² - (2x - 5) (3x + 7)
    C = (2 * 1/2 - 5)² - (2 * 1/2 - 5) (3 * 1/2 + 7)
    C = (2/2 - 5)² - (2/2 - 5) (3/2 + 7 x 2 / 1 x 2)
    C = - 4² - (2/2 - 5 x 2 / 1 x 2) (3/2 +14/2)
    C = 16 - (2/2 - 10/2) (17/2)
    C = 16 - (- 8/2) (17/2)
    C = 16 - (- 136/4)
    C = 16 x 4 / 1 x 4 + 152/4
    C = 64/4 + 136/4
    C = 200/4
    C = 50

    Développer et réduire C
    C = (2x - 5)² - (2x - 5) (3x + 7)
    C = 4x² - 20x + 25 - (6x² + 14x - 15x - 35)
    C = 4x² - 20x + 25 - 6x² - 14x + 15x + 35
    C = 4x² - 6x² - 20x - 14x + 15x + 25 + 35
    C = - 2x² - 19x + 60



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