题目

计算[x+(1)/x][x^2+(1)/x^2][x^4+(1)/x^4][x^8+(1)/x^8][x^16+(1)/x^16](x^2-1) 主要是解题

提问时间：2020-11-13

答案

[x+(1)/x][x^2+(1)/x^2][x^4+(1)/x^4][x^8+(1)/x^8][x^16+(1)/x^16](x^2-1)
=[x-(1)/x][x+(1)/x][x^2+(1)/x^2][x^4+(1)/x^4][x^8+(1)/x^8][x^16+(1)/x^16](x^2-1)/[x-(1)/x]
=[x^2-(1)/x^2][x^2+(1)/x^2][x^4+(1)/x^4][x^8+(1)/x^8][x^16+(1)/x^16](x^2-1)/[(x^2-1)/x]
=[x^4-(1)/x^4][x^4+(1)/x^4][x^8+(1)/x^8][x^16+(1)/x^16](x^2-1)* [x/(x^2-1)]
=[x^8-(1)/x^8][x^8+(1)/x^8][x^16+(1)/x^16]*x
=[x^16-(1)/x^16][x^16+(1)/x^16]*x
=x[x^32-(1)/x^32]