初三数学-5-业主生活–手机房天下问答

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老卒B站丨Lv 1

设x^2-2x-1=0的两根分别为m,n.则：m+n=2;mn=-1.【只要求得(m^7+n^7)和(m^7*n^7)的值即可得方程】1)(m+n)^2=4,m^2+n^2+2mn=4,m^2+n^2=6; (m^2+n^2)^2=36,m^4+n^4+2(mn)^2=36,m^4+n^4=34; (m+n)*(m^2+n^2)=2*6,m^3+n^3+mn(m+n)=12,m^3+n^3=14.∴(m^3+n^3)(m^4+n^4)=14*34, m^7+n^7+(mn)^3*(m+n)=476, m^7+n^7=478.2)m^7*n^7=(mn)^7=-1.故所求方程为：x^2-478x-1=0.

+1 2011-01-11 举报