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  • 在有理數(shù)范圍內(nèi)因式分解: (1)16(6x-1)(2x-1)(3x+1)(x-1)+25=_. (2)(6x-1)(2x-1)(3x-1)(x-1)+x2=_. (3)(6x-1)(4x-1)(3x-1)(x-1)+9x4=_.

    在有理數(shù)范圍內(nèi)因式分解:
    (1)16(6x-1)(2x-1)(3x+1)(x-1)+25=______.
    (2)(6x-1)(2x-1)(3x-1)(x-1)+x2=______.
    (3)(6x-1)(4x-1)(3x-1)(x-1)+9x4=______.
    數(shù)學(xué)人氣:143 ℃時(shí)間:2019-10-14 00:52:20
    優(yōu)質(zhì)解答
    (1)16(6x-1)(2x-1)(3x+1)(x-1)+25,
    =[(6x-1)(4x-2)][(6x+2)(4x-4)]+25,
    =(24x2-16x+2)(24x2-16x-8)+25,
    =(24x2-16x)2-6(24x2-16x)-16+25,
    =(24x2-16x)2-6(24x2-16x)+9,
    =(24x2-16x-3)2;
    (2)(6x-1)(2x-1)(3x-1)(x-1)+x2,
    =[(6x-1)(x-1)][(2x-1)(3x-1)]+x2,
    =(6x2-7x+1)(6x2-5x+1)+x2,
    =(6x2-6x+1-x)(6x2-6x+1+x)+x2,
    =(6x2-6x+1)2-x2+x2,
    =(6x2-6x+1)2
    (3)(6x-1)(4x-1)(3x-1)(x-1)+9x4,
    =[(6x-1)(x-1)][(4x-1)(3x-1)]+9x4,
    =(6x2-7x+1)(12x2-7x+1)+9x4
    令t=6x2-7x+1,則12x2-7x+1=t+6x2,
    ∴原式=t(t+6x2)+9x4
    =t2+6?t?x2+9x4,
    =(t+3x22
    =(6x2-7x+1+3x22,
    =(9x2-7x+1)2
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