Question

高一数学题，求解，谢谢

Answer 1

你好，我是杭州庆春路精锐教育的老师，已知 x+(1/x)=a,则 [x+(1/x)]²=a² → x²+(1/x²)=a²-2
[x+(1/x)]*[x²+(1/x²)]=x³+(1/x³)+[x+(1/x)]=a*(a²-2) → x³+(1/x³)=a³-3a；
[x²+(1/x²)]*[x³+(1/x³)]=[x^5+(1/x^5)]+[x+(1/x)]=(a²-2)*(a³-3a)
→ x^5+(1/x^5)=(a²-2)(a³-3a)-a；
[x³+(1/x³)]*[x^5+(1/x^5)]=[x^8+(1/x^8)]+[x²+(1/x²)]=(a³-3a)[(a²-2)(a³-3a)-a]
→ x^8+(1/x^8)=(a³-3a)[(a²-2)(a³-3a)-a]-(a²-2)；
[x^5+(1/x^5)]*[x^8+(1/x^8)]=[x^13+(1/x^13)]+[x³+(1/x³)]
=[(a²-2)(a³-3a)-a]*{(a³-3a)[(a²-2)(a³-3a)-a]-(a²-2)}
∴ x^13+(1/x^13)=[(a²-2)(a³-3a)-a]*{(a³-3a)[(a²-2)(a³-3a)-a]-(a²-2)}-(a³-3a)；

追问

谢谢