所以a≥1.

答案:[1,+∞)

6.已知关于x的不等式|x-2|0)的解集为{x|-1

解析:不等式|x-2|

　　所以不等式的解集为{x|2-a

　　由题意,知{x|2-a

　　则{■(2"-" a="-" 1"," @2+a=b"," )┤解得{■(a=3"," @b=5"." )┤

　　故a+2b=3+10=13.

答案:13

7.若关于x的不等式|2x-1|+|x+2|≥a2+1/2a+2对任意实数x恒成立,则实数a的取值范围是　　　　　.

解析:令f(x)=|2x-1|+|x+2|={■("-" 3x"-" 1"," x≤"-" 2"," @3"-" x",-" 21/2 "," )┤可求得f(x)的最小值为5/2,故原不等式恒成立可转化为a2+1/2a+2≤5/2恒成立,即a2+a/2-1/2≤0,即(a+1)(a"-" 1/2)≤0,解得a∈["-" 1"," 1/2].

答案:["-" 1"," 1/2]

8.若关于x的不等式|3x-b|<4的整数解有且仅有1,2,3,则b的取值范围是　　　　　　　.

解析:由|3x-b|<4,得-4<3x-b<4,

　　即("-" 4+b)/3

　　∵不等式|3x-b|<4的整数解有且仅有1,2,3,则{■(0≤("-" 4+b)/3<1"," @3<(4+b)/3≤4"," )┤

　　即{■(4≤b<7"," @5

答案:(5,7)

9.解下列不等式:

(1)3≤|x-2|<9;(2)|3x-4|>1+2x.

解(1)原不等式等价于不等式组{■("|" x"-" 2"|" ≥3"," @"|" x"-" 2"|" <9"." )┤0├ ■("①" @"②" )┤

　　由①,解得x≤-1或x≥5.

由②,解得-7