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  • (x-1)(x+3)=12,(3x-5)(2x-1)= -12x+7,(6x+5)(6y-5)-24=0,4(x-3)²-x(x-3)=0用因式分解怎么解?

    (x-1)(x+3)=12,(3x-5)(2x-1)= -12x+7,(6x+5)(6y-5)-24=0,4(x-3)²-x(x-3)=0用因式分解怎么解?
    數(shù)學(xué)人氣:582 ℃時間:2020-04-01 05:03:23
    優(yōu)質(zhì)解答
    (x-1)(x+3)=12
    x^2 + 2x - 3 = 12 ,x^2 + 2x - 15 = 0 = (x+5)(x-3)
    x = 5 ,x = 3
    (3x-5)(2x-1)= -12x+7 = 6x^2 - 13x + 5
    6x^2 - x - 2 = 0 = (2x + 1)(3x - 2)
    x = -1/2 ,x = 2/3
    (6x+5)(6x-5)-24=0 = 36x^2 - 49 = (6x + 7)(6x - 7)
    x = -7/6 ,x = 7/6
    4(x-3)2-x(x-3)=0 = (x - 3)[4(x - 3) - x] = 3(x - 3)(x - 4)
    x = 3 ,x = 4
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