Question

设x-1\x=3,求x^10+x^8+x^2+1\x^10+x^6+x^4+1的值,

Answer 1

x-1/x=3

=>x^2+1/x^2=(x-1/x)^2+2=3^2+2=11

(x+1/x)^2=(x-1/x)^2+4=9+4=13

=>x+1/x=√13

=>x^3+1/x^3=(x+1/x)(x^2+1/x^2-1）=√13*（11-1）=10√3

=>x^4+1/x^4=（x^2+1/x^2)^2-2=11^2-2=119

(x^10+x^8+x^2+1)/x^10+x^6+x^4+1

=(x^6(x^4+1/x^4)+x^4(x^4+1/x^4))/(x^7（x^3+1/x^3)+x^3(x^3+1/x^3))

=119(x^6+x^4)/(10√3*(x^7+x^3))

=119x^5(x+1/x)/(10√3x^5(x^2+1/x^2))

=119*3/(10√3*11)

=119√3/110