Mathématiques

Question

SVP Urgent ! Exercice de DM: Résoudre les équations suivantes

1)[tex] x^{2} + 13=48[/tex]
2)[tex]7 x^{2} -11=3 x^{2} +13[/tex]
3)[tex](5x-2) ^{2} =36[/tex]
4)[tex](3x+2) ^{2} = (5x-4) ^{2} [/tex]

1 Réponse

  • Salut,

    x² + 13 = 48
    x² = 48 - 13
    x² = 35
    x² - 35 = 0
    (x+[tex] \sqrt{35} [/tex])(x-[tex] \sqrt{35} [/tex]) = 0
    x - [tex] \sqrt{35} [/tex] = 0
    x = [tex] \sqrt{35} [/tex]
    x + [tex] \sqrt{35} [/tex] = 0
    x = -[tex] \sqrt{35} [/tex]

    7x² - 11 = 3x² + 13
    7x² -3x² = 13 +11
    4x² = 24
    4x² - 24 = 0
    (2x - [tex] \sqrt{24} [/tex])(2x+[tex] \sqrt{24} [/tex]) = 0
    2x - [tex] \sqrt{24} [/tex] = 0
    2x = [tex] \sqrt{24} [/tex]
    x = [tex] \sqrt{6} [/tex]
    2x+[tex] \sqrt{24} [/tex] = 0
    2x = -[tex] \sqrt{24} [/tex]
    x = -[tex] \sqrt{6} [/tex]

    (5x-2)² = 36
    (5x-2)² - 36 = 0
    (5x-2+6)(5x-2-6) = 0
    (5x +4)(5x-8) = 0
    5x + 4 = 0
    5x = -4
    x = -4/5

    5x - 8 = 0
    5x = 8
    x = 8/5

    (3x+2)² = (5x-4)²
    (3x+2)² - (5x-4)² = 0
    [(3x + 2) - (5x-4)][(3x+2)+(5x-4)] = 0
    (3x+2-5x+4)(3x+2+5x-4) = 0
    (8x + 6)(8x -2) = 0
    2(4x+3)(8x-2) = 0
    4(4x+3)(4x-1) = 0
    4x + 3 = 0
    4x = -3
    x = -3/4
    4x - 1 = 0
    4x = 1
    x = 1/4

    Bonne soirée !












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