一道初二较难的数学题

来源：百度知道 编辑：UC知道 时间：2024/05/14 19:41:39
设x<0且x- 1/x =根号5 ,求 (x10+x6+x4+1)/(x10+x8+x2+1) 的值
x10是x的10次方，x6是x的6次方……依此类推，1为常数项

x- 1/x =√5
平方
x^2-2+1/x^2=5
x^2+1/x^2=7
(x+1/x)^2=x^2+2+1/x^2=9
x<0
x+1/x<0
x+1/x=-3
x^3+1/x^3
=(x+1/x)(x^2-x*1/x+1/x^2)
=(-3)*(7-1)
=-18
x^5+1/x^5
=(x^3+1/x^3)(x^2+1/x^2)-x^3*1/x^2-x^2*1/x^3
=(-18)*7-(x+1/x)
=-126+3
=-123

(x10+x6+x4+1)/(x10+x8+x2+1)
上下同除以x^5
=[(x^5+1/x^5)+(x+1/x)]/[(x^5+1/x^5)+(x^3+1/x^3)]
=(-123-3)/(-123-18)
=42/47

把x- 1/x =根号5两边平方,得x^2+1/x^2=7,在平方得x^4+1/x^4=47