问题标题：
一道数学题,希望能用简便的方法解答.设x-1/x=3,求x十次方+x八次方+x平方+1除以x十次方+x六次方+x四次方+1的值
问题描述：
一道数学题,希望能用简便的方法解答
.设x-1/x=3,求x十次方+x八次方+x平方+1除以x十次方+x六次方+x四次方+1的值
黄劲回答：
已知：x-1/x=3,求：(x^10+x^8+x^+1)/(x^10+x^6+x^4+1)的值
x-1/x=3
(x-1/x)^2=x^2+1/x^2-2=9
x^2+1/x^2=11
(x+1/x)^2=x^+2+1/x^2=13
x+1/x=t,则t^2=13
x^3+1/x^3
=(x+1/x)(x^2-1+1/x^2)
=(x+1/x)[(x^2+2+1/x^2)-3]
=(x+1/x)[(x+1/x)^2-3]
=t(13-3)
=10t
x^5+1/x^5
=(x^2+1/x^2)(x^3+1/x^3)-(x+1/x)
=11*10t-t
=109t
(x^10+x^8+x^+1)/(x^10+x^6+x^4+1).上下同除以x^5
=[(x^5+1/x^5)+(x^3+1/x^3)]/[(x^5+1/x^5)+(x+1/x)]
=[109t+10t]/[109t+t]
=119/110
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