设x＜0,且x减x分之一等于根号五,求有理式x10+x6+x4+1除以x10+x8+x2+1的值

Question

设x＜0,且x减x分之一等于根号五,求有理式x10+x6+x4+1除以x10+x8+x2+1的值

Answer 1

x-1/x=根号5
(x-1/x)^2=5,x^2+1/x^2-2=5,x^2+1/x^2=7
(x^2+1/x^2)^2=49,x^4+1/x^4+2=49,x^4+1/x^4=47
(x^10+x^6+x^4+1)/(x^10+x^8+x^2+1)
=[x^2(x^8+x^4+x^2+1/x^2)]/[x^4(x^6+x^4+1/x^2+1/x^4)]
=[(x^8+x^4)+(x^2+1/x^2)]/[x^2(x^6+1/x^2)+(x^4+1/x^4)]
=[x^6(x^2+1/x^2)+(x^2+1/x^2)]/{x^2[(x^6+1/x^2)+(x^4+1/x^4)]}
=[(x^6+1)(x^2+1/x^2)] / {x^2[x^2(x^4+1/x^4)+(x^4+1/x^4)]}
=[(x^6+1)*7] / [x^2(x^2+1)(x^4+1/x^4)]
=[(x^6+1)*7] / [x^2(x^2+1)*47]
=[7(x^2+1)(x^4-x^2+1)] / [47x^2(x^2+1)]
=[7(x^4-x^2+1)] / [47x^2]
=7(x^2-1+1/x^2)/47
=7(x^2+1/x^2-1)/47
=7(7-1)/47
=42/47