x+1/x=-1,则x^2005+1/x^2005=______

Question

要有计算过程

Answer 1

楼上二位都是错的。正确答案是：-1。

x^2005+1/x^2005
=(x+1/x)(x^2004+1/x^2004)-(x^2003+1/x^2003)
=-(x^2004+1/x^2004)-(x^2003+1/x^2003)
=-[(x+1/x)(x^2003+1/x^2003)-(x^2002+1/x^200)]-(x^2003+1/x^2003)
=-[-(x^2003+1/x^2003)-(x^2002+1/x^200)]-(x^2003+1/x^2003)
=x^2002+1/x^2002
............................（同理可推）
=x+1/x （因为2005=668*3+1）
=-1。

为什么二楼是错的？

因为
x3+1/x^3
=(x+1/x)(x^2+1/x^2)-(x+1/x)
=-(x^2+1/x^2)-(x+1/x)
=-[(x+1/x)(x+1/x)-2]-(x+1/x)
=-[(-1)*(-1)-2]-(-1)
=2。（不等于-1）
其用数学归纳法，错在（3）步：
(3)，(x+1/x)(x^k+1/x^k)
=x^(k+1)+1/x^(k+1)+2
[这里不成立，应为x^(k+1)+1/x^(k+1)+x^(k-1)+1/x^(k-1) ]

Answer 2

x^2005+1/x^2005=-1

用数学归纳法 我们证明x^n+1/x^n=-1，n∈N*
(1)n=1时结论成立
(2)假设对于n=k,k>=1,结论成立，即x^k+1/x^k=-1
(3)有(x+1/x)(x^k+1/x^k)=1
=x^(k+1)+1/x^(k+1)+2
x^(k+1)+1/x^(k+1)=-1成立
即对n=k+1结论成立
所以有对一切的n∈N*,x^n+1/x^n=-1成立
n=2005时,即有x^2005+1/x^2005=-1

Answer 3

x+1/x=-1
=>x^2+1=-x
=>x^2+x+1=0
=>x≠1
（x^2+x+1）（x-1）=0
=>x^3-1=0
=>x^3=1
=>x^2005=x
∴x^2005+1/x^2005=x+1/x=-1

Answer 4

x=-1/2;
代进去之后原式=-2^2005+1!