在有理数范围内因式分解:
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在有理数范围内因式分解:
(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=______.
(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=______.
(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,则12x2-7x+1=t+6x2,
∴原式=t(t+6x2)+9x4,
=t2+6•t•x2+9x4,
=(t+3x2)2,
=(6x2-7x+1+3x2)2,
=(9x2-7x+1)2.
=[(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,则12x2-7x+1=t+6x2,
∴原式=t(t+6x2)+9x4,
=t2+6•t•x2+9x4,
=(t+3x2)2,
=(6x2-7x+1+3x2)2,
=(9x2-7x+1)2.
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