题目内容

已知x+
1
x
=2
,则下列等式成立的有(  )
x2+
1
x2
=2
;②x4+
1
x4
=2
;③x8+
1
x8
=2
;④x-
1
x
=0
分析:根据完全平方公式求出x2+
1
x2
=(x+
1
x
2-2•x•
1
x
=22-2=2,x4+
1
x4
=(x2+
1
x2
2-2•x2
1
x2
=22-2=2,x8+
1
x8
=(x4+
1
x4
2-2=2,(x-
1
x
2=(x+
1
x
2-4•x•
1
x
=22-4=0,判断即可.
解答:解:∵x+
1
x
=2,
∴x2+
1
x2
=(x+
1
x
2-2•x•
1
x
=22-2=2,∴①正确;
x4+
1
x4
=(x2+
1
x2
2-2•x2
1
x2
=22-2=2,∴②正确;
x8+
1
x8
=(x4+
1
x4
2-2=2,∴③正确
(x-
1
x
2=(x+
1
x
2-4•x•
1
x
=22-4=0,∴④正确;
故选D.
点评:本题考查了完全平方公式的应用,注意:(a+b)2=a2+2ab+b2,(a-b)2=a2-2ab+b2
练习册系列答案
相关题目

违法和不良信息举报电话:027-86699610 举报邮箱:58377363@163.com

精英家教网