Question

计算:(x-1/2)(2x+1)(2x^2+1/2)/(16x^4-1)

Answer 1

解：最后一个应该是加吧，如果是＋，就这样计算，如果不是也容易。
(x-1/2)(2x+1)(2x^2+1/2)/(16x^4+1)
＝1／4×(2x-1)(2x+1)(4x^2+1)(16x^4＋1)
＝1／4×(4x^2-1)(4x^2+1)(16x^4＋1)
＝1／4×(16x^4-1)(16x^4＋1)
＝1／4×(256x^8-1)
＝64x^8-1／4

Answer 2

原式=(x-1/2)(2x+1)(2x^2+1/2)/(4x^2+1）（4x^2-1)
=(x-1/2)(2x+1)(2x^2+1/2)/（2x+1）（2x-1）(4x^2+1)
=(x-1/2)(2x^2+1/2)/(2x-1)(4x^2+1)
=1/4

Answer 3

1/4 ， 原式=(x-1/2)(2x+1)(2x^2+1/2)/(4x^2+1）（4x^2-1)
=(x-1/2)(2x+1)(2x^2+1/2)/（2x+1）（2x-1）(4x^2+1)
=(x-1/2)(2x^2+1/2)/(2x-1)(4x^2+1)
=1/4

Answer 4

^是什么意思啊

Answer 5

是简化么，我的答案是：（（2x +1）(4x ^2 +1)(x - 1/2))^2