已知x+1/x=2,求x2048+x−2048−2x2013+x−2013.

问题描述:

已知x+

1
x
=2,求
x2048+x−2048−2
x2013+x−2013

∵x+

1
x
=2,
∴(x+
1
x
2=4,
∴x2+
1
x2
=2,
同理可得,x4+
1
x4
=2…x2048+
1
x2048
=2,
∴原式=
(x2048+
1
x2048
)−2
x2013+x−2013
=
2−2
x2013+x−2013
=0.