计算:(x+2)(4x-2)-(2x-1)(2x+1)-(x-4)的平方 求大神····
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计算:(x+2)(4x-2)-(2x-1)(2x+1)-(x-4)的平方 求大神····
计算:(x+2)(4x-2)-(2x-1)(2x+1)-(x-4)的平方 求大神····
计算:(x+2)(4x-2)-(2x-1)(2x+1)-(x-4)的平方 求大神····
(x+2)(4x-2)-(2x-1)(2x+1)-(x-4)^2
=(2x+4)(2x-1)-(2x-1)(2x+1)-(x-4)^2
=(2x+4-2x-1)(2x-1)-(x-4)^2
=3(2x-1)-(x-4)^2
=6x-3-(x^2-8x+16)
=6x-3-x^2+8x-16
=-x^2+14x-19.你看对不对.
(x+4)^2(x-4)^2 - (2x-1)^2(2x+1)^2
= 【(x+4)(x-4)】^2 - 【(2x-1)(2x+1)】^2
= (x^2-16)^2 - (4x^2-1)^2
= x^4-32x^2+256 - (16x^4-8x^2+1)
= -15x^4-24x^2+255
4x^2-2x+8x-4-(4x^2-1)-x^2+8x-16
=-x^2+14x-19
原式=(4X²-2X+8X-4)-(4X²-1)-(X²-8X+16)
=-X²+14X-19
先把它们的括号都打开↓
原式=4x方+6x-4-4x方+1-x方+8x-16
再将能约分的约掉↓,合并相同的,则
=-x方+14x-19
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