Question

x+1/x=3,x的五次方加上x的负五次方值为多少?求具体过程

Answer 1

已知：x+1/x=3，那么：(x+1/x)²=9
即 x² +2*x*(1/x)+ 1/x²=9
可得： x² + 1/x²=7
那么：x³ + 1/x³=(x + 1/x)( x² - 1 + 1/x²)=3*8=24
所以：( x² + 1/x²)(x³ + 1/x³)
=x的5次幂 + x²*(1/x³) + (1/x²)*x³ + 1/x的5次幂
=x的5次幂 + x的-5次幂 + 1/x +x
=7*24
=168
即得：x的5次幂 + x的-5次幂=168-(1/x +x)=168-3=165

Answer 2

x+1/x=3
则：x²+1/x²=7
上式等号两边同时平方得：x^4+1/x^4=47
那么：x^5+1/x^5
=(x+1/x)[(x^4+1/x^4)-(x²+1/x²)+1]
=3*(47-7+1)
=123

Answer 3

x^2+1/x^2=(x+1/x)^2-2=7
x^3+1/x^3=(x^2+1/x^2)(x+1/x)-(x+1/x)=18
x^4+1/x^4=(x^2+1/x^2)^2-2=47
x^5+1/x^5=(x^4+1/x^4)(x+1/x)-(x^3+1/x^3)=47*3-18=123