(1) |2x+1|+|x-2|>4
(i)x -(2x+1)+2-x>4
=>x14
=>x>5/3
=>x>2 (c)
综合(a)(b)(c)=>x1
因此x∈（-∞,-1）∪（1,∞）
（2)
|x^2+2|>3|x|
=>x^2+2>3|x|
（i)当x>=0时
=>x^2+2>3x
=>x^2-3x+2>0
=>x>2或x0x^2+3x+2>0
=>x>-1或x-1