本善良友爱
幼苗
共回答了14个问题采纳率:100% 举报
将已知两边取倒数,得:
(x^2-mx+1)/x=1
x-m+1/x=1
x+1/x=1+m
则:
x^3+1/x^3
=(x+1/x)(x^2-1+1/x^2)
=(x+1/x)[(x+1/x)^2-3]
=(1+m)[(1+m)^2-3]
=(1+m)(m^2+2m-2)
=(m^2+2m-2)+m(m^2+2m-2)
=m^3+3m^2-2
先算:
(x^6-m^3x^3+1)/x^3
=x^3-m^3+1/x^3
=x^3+1/x^3-m^3
=m^3+3m^2-2-m^3
=3m^2-2
所以
x^3/(x^6-m^3x^3+1)=1/(3m^2-2)
故应选C.
1年前
1