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=_.
问题描述:
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=______.
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...