计算:(x+1)(x的平方+）（x的2次方+1）（8次方+1)(x的16次方+1)(x的32次方+1)_数学解答

计算:(x+1)(x的平方+）（x的2次方+1）（8次方+1)(x的16次方+1)(x的32次方+1)

人气：256 ℃　时间：2019-11-23 13:25:53

解答

如果x=1,则原式=2^6=64
如果x不等于1,则
原式=(x-1)(x+1)(x^2+1)(x^4+1)(x^8+1)(x^16+1)(x^32+1)/(x-1)
=(x^2-1)(x^2+1)(x^4+1)(x^8+1)(x^16+1)(x^32+1)/(x-1)
=(x^4-1)(x^4+1)(x^8+1)(x^16+1)(x^32+1)/(x-1)
=(x^8-1)(x^8+1)(x^16+1)(x^32+1)/(x-1)
=(x^16-1)(x^16+1)(x^32+1)/(x-1)
=(x^32-1)(x^32+1)/(x-1)
=(x^64-1)/(x-1)