![](http://thumb.1010pic.com/pic5/upload/201310/5283bccff0b0b.png)
解:(1)实数与数轴上的点是一一对应关系,故本小题错误;
(2)①底数是1时,x
2-x-1=1,
解得x=2或x=-1,
②底数是-1时,指数必须为偶数,
x
2-x-1=-1且x+2为偶数,
解得x=0或x=1(舍去),
③底数不等于0,指数为0时,
x+2=0,
解得x=-2,
此时,底数x
2-x-1=4-(-2)-1=1≠0,
综上所述x为-1、2、0、-2,故本小题错误;
(3)如图,∠1与∠2、∠3的两边分别垂直,∠1与∠2互补,∠1与∠3相等,故本小题正确;
(4)若圆的半径为5,AB、CD是两条平行弦,且AB=8,CD=6,
根据垂径定理,AB、CD的弦心距分别为:
![](http://thumb.1010pic.com/pic5/latex/459636.png)
=3,
![](http://thumb.1010pic.com/pic5/latex/38804.png)
=4,
![](http://thumb.1010pic.com/pic5/upload/201310/5283bcd0100fa.png)
如图①,AC=
![](http://thumb.1010pic.com/pic5/latex/498544.png)
=
![](http://thumb.1010pic.com/pic5/latex/53.png)
,
或AC=
![](http://thumb.1010pic.com/pic5/latex/498545.png)
=5
![](http://thumb.1010pic.com/pic5/latex/53.png)
,
如图②,AC=
![](http://thumb.1010pic.com/pic5/latex/498546.png)
=5
![](http://thumb.1010pic.com/pic5/latex/53.png)
,
或AC=
![](http://thumb.1010pic.com/pic5/latex/498547.png)
=7
![](http://thumb.1010pic.com/pic5/latex/53.png)
,
综上所述,弦AC的长为
![](http://thumb.1010pic.com/pic5/latex/53.png)
或5
![](http://thumb.1010pic.com/pic5/latex/53.png)
或7
![](http://thumb.1010pic.com/pic5/latex/53.png)
,故本小题错误;
(5)设A、B两点坐标为(x
1,0)(x
2,0),
则AB=|x
1-x
2|=
![](http://thumb.1010pic.com/pic5/latex/326041.png)
=
![](http://thumb.1010pic.com/pic5/latex/498548.png)
,
顶点P的纵坐标为
![](http://thumb.1010pic.com/pic5/latex/8487.png)
=
![](http://thumb.1010pic.com/pic5/latex/498549.png)
,
∵△PAB是正三角形,
∴
![](http://thumb.1010pic.com/pic5/latex/376.png)
×
![](http://thumb.1010pic.com/pic5/latex/498548.png)
=-
![](http://thumb.1010pic.com/pic5/latex/498549.png)
,
设m=b
2-16,
则方程可化为m
2-12m=0,
解得m
1=0(舍去),m
2=12,
即b
2-16=12,
解得b=±2
![](http://thumb.1010pic.com/pic5/latex/4546.png)
,故本小题错误;
综上所述,(1)(2)(4)(5)错误.
故答案为:(1)(2)(4)(5).
分析:(1)根据数轴上的点与实数的关系解答;
(2)分①底数是1,②底数是-1,而指数是偶数,③底数不等于0,而指数为0三种情况分别进行时求解;
(3)作出图形,利用数形结合验证;
(4)根据垂径定理,以及等腰梯形的性质,分AB、CD在圆心的同侧与异侧,并AC是相邻的两点与不相邻的两点共四种情况讨论求解;
(5)利用根与系数的关系求出AB的长,利用顶点坐标求出点P的纵坐标,然后根据等边三角形边长与高的关系列出方程,然后解方程即可.
点评:本题综合考查了实数与数轴的对应关系,任何非0数的0次幂等于1,垂径定理以及抛物线与x轴的交点问题,等边三角形的性质,本题特点在于分情况讨论,不要漏解而导致出错.