金庸群侠传3 狄云:若m^2=m+1,n^2=n+1,且m 不等于n,则m^5+n^5=?
来源:百度文库 编辑:高考问答 时间:2024/05/13 14:44:37
很难
不停的利用m+1来替换m^2:
m^5=m*(m^2)^2=m(m+1)^2=m(m^2+2m+1)=m(m+1+2m+1)=m(3m+2)=3m^2+2m=3(m+1)+2m=5m+3;
同理,n^5=5n+3
又m^2=m+1,n^2=n+1;把两式相减的m^2-n^2=m-n,即(m+n)(m-n)=m-n
因为m不等于n,两边除以m-n得m+n=1
所以m^5+n^5=5m+3+5n+3=5(m+n)+6=5+6=11
11
m和n是方程x^2-x-1=0的两个根,
m+n=1
mn=-1
m^2+n^2=m+1+n+1=3
m^2 * n^2=(m+1)(n+1)=m+n+mn+1=1
m^3+n^3=(m+n)(m^2-mn+n^2)=2
(m^3+n^3)(m^2+n^2)=m^5+n^5+m^3 * n^2+m^2 * n^3
m^5+n^5=(m^3+n^3)(m^2+n^2)-m^2 * n^2 (m+n)=5
思路基本是这样,得数不一定对,自己再算算吧
m^2=m+1 (1)
n^2=n+1 (2)
(1)-(2)得m+n=1
m^5+n^5
=m*m^2*m^2+n*n^2*n^2
=m*(m+1)^2+n*(n+1)^2
=m^3+2m^2+m+n^3+2n^2+n
=m(m+1)+2m^2+m+n(n+1)+2n^2+n
=3m^2+2m+3n^2+2n
=5m+3+5n+3
=5(m+n)+6
=11
1楼正解
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