当x-1/x=3时,(x10+x8+x2+1)/(x10+x6+x4+1)=

Question

当x-1/x=3时,(x10+x8+x2+1)/(x10+x6+x4+1)=

Answer 1

首先你得知道立方和公式：a^3+b^3=(a+b)(a^2-ab+b^2)
由x-1/x=3可得(x-1/x)^2=9,即x^2+1/x^2=11,再平方可得x^4+1/x^4=119.
所以所求式子
=(x^5+x^3+1/x^3+1/x^5)/(x^5+x^4+1/x^4+1/x^5)
=(x+1/x)(x^4+1/x^4)/(x^2+1/x^2)(x^3+1/x^3)
利用立方和公式可知：x^3+1/x^3=(x+1/x)(x^2-1+1/x^2)
所以所求式子
=(x+1/x)(x^4+1/x^4)/(x^2+1/x^2)(x^2-1+1/x^2)(x+1/x)
=(x^4+1/x^4)/(x^2+1/x^2)(x^2-1+1/x^2)
=119/[(11-1)*10]
=119/110
检验一下看看