Question

若不等式组{x-a>2,b-2x>0的解集式-1<x<1,则(a+b)的2008次方=

Answer 1

x-a>2,b-2x>0
=>a+2<x<b/2.
不等式组{x-a>2,b-2x>0的解集式-1<x<1.
=>a+2=-1;b/2=1.
<=>a=-3;b=2.
=>a+b=-3+2=-1.
=>(a+b)的2008次方=1.

Answer 2

x-a>2
a>a+2
b-2x>0
2x<b
x<b/2
所以a+2<x<b/2
解集是-1<x<1,
所以a+2=-1,b/2=1
a=-3,b=2

(a+b)^2008
=(-1)^2008
=1

Answer 3

x-a>2，x>a+2
b-2x>0,2x<b,x<b/2
即a+2<x<b/2
解集式-1<x<1
a+2=-1,a=-3
b/2=1,b=2
(a+b)^2008
=(-3+2)^2008
=1^2008
=1

Answer 4

a+2<x<b/2

a+2=-1
b/2=1
a=-3,b=2
a+b=-1
(a+b)^2008=1

Answer 5

x＞a+2 a=-3
2x＜b b=2
(a+b)^2008=1