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  • 1 (x²+3x-3)(x²+3x+4)-8=

    1 (x²+3x-3)(x²+3x+4)-8=
    2 (x-1)x(x+1)(x+2)-24=
    3 2x²+xy-y²-4x+5y-6=
    數(shù)學(xué)人氣:177 ℃時(shí)間:2020-02-03 00:18:25
    優(yōu)質(zhì)解答
    (x²+3x-3)(x²+3x+4)-8
    =(x^2+3x)^2+(x^2+3x)-20
    =(x^2+3x-4)(x^2+3x+5)
    =(x-1)(x+4)(x^2+3x+5)
    x(x + 1)(x - 1)(x + 2) - 24
    = x(x^2 - 1)(x + 2) - 24
    = x(x^3 + 2x^2 - x - 2) - 24
    = x^4 + 2x^3 - x^2 - 2x - 24
    = (x^4 - 2x^3) + 4x^3 - x^2 - 2x - 24
    = x^3(x - 2) + 4x^3 - 8x^2 + 7x^2 - 2x - 24
    = (x^3 + 4x^2)(x - 2) + (7x^2 - 14x) + (12x - 24)
    = (x^3 + 4x^2 + 7x + 12)(x - 2)
    = [(x^3 + 3x^2) + (x^2 + 3x) + (4x + 12)](x - 2)
    = (x^2 + x + 4)(x + 3)(x - 2)
    2x²+xy-y²-4x+5y-6
    =2x^2+(y-4)x+(-y^2+5y-6)
    =2x^2+(y-4)x+[-(y^2-5y+6)]
    =2x^2+(y-4)x+[-(y-2)(y-3)]
    =[1x+(y-3)][2x+(-y+2)]
    =(x+y-3)(2x-y+2)
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