Mathématiques

Question

II Développer et reduire
H=3(4a-3) I=(y+2)(y+5)
J=(4c-3)(2c+1) K=(x-4)(2x-3)

2 Réponse

  • H= 3(4a - 3)
    H= 12a - 9

    I= (y + 2)(y + 5)
    I= y² + 5y + 2y + 10
    I= y² + 7y + 10

    J= (4c - 3)(2c + 1)
    J= 8c² + 4c - 6c - 3
    J= 8c² - 2c - 3

    K= (x - 4)(2x - 3)
    K= 2x² - 3x - 8x + 12
    K= 2x² - 11x + 12
  • H=3(4a-3)
    = 12a-9

     I=(y+2)(y+5)

    = y^2+5y+2y+10
    = y^2+ 7y + 10

    J=(4c-3)(2c+1)

    = 8c^2+4c-6c-3
    = 8c^2 -2c -3

    K=(x-4)(2x-3)
    = 2x^2-3x-8x+12
    = 2x^2 -11x + 12






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