Mathématiques

Question

Developper puis reduire

(2x+1)(x-4)
(x-3)(2x-3)
(x+2)-(2x-1)
(2x-3)(3x+5)
(x+3)(2-4x)

Mercii

2 Réponse

  • Développer puis réduire
    (2x+1)(x-4)
    = 2x² - 8x + x - 4
    = 2x² - 7x - 4

    (x-3)(2x-3)
    = 2x² - 3x - 6x + 9
    = 2x² - 9x + 9

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

    (2x-3)(3x+5)
    = 6x² + 10x - 9x - 15
    = 6x² + x - 15

    (x+3)(2-4x)

    = 2x - 4x² + 6 - 12x
    = - 4x² - 10x + 6
  • = ( 2x + 1 ) ( x - 4 )
    = 2x * x + 2x * ( - 4 ) + 1 * x + 1 * ( - 4 )
    = 2x² - 8x + x - 4
    = 2x² - 7x - 4 

    = ( x - 3 ) ( 2x - 3  )
    = x * 2x + x * ( - 3 ) - 3 * 2x - 3 * ( - 3 )
    = 2x² - 3x - 6x + 9
    = 2x² - 9x + 9

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

    = ( 2x - 3 ) ( 3x + 5 )
    = 2x * 3x + 2x * 5 - 3 * 3x - 3 * 5
    = 6x² + 10x - 9x - 15
    = 6x² + x - 15

    = ( x + 3 ) ( 2 - 4x )
    = x * 2 + x * ( - 4x ) + 3 * 2 + 3 * ( - 4x )
    = 2x - 4x² + 6 - 12x
    = - 4x² - 10x + 6

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