Question

高一数学解不等式

(1) |2x+1|+|x-2|>4
(2) |x^2+2|>3|x|
(3) 解不等式组: (x^2+1)(x-3)≤0
|3x-4|≤0

急~~~~ 谢了

Answer 1

(1) |2x+1|+|x-2|>4
(i)x<=-1/2时
原式=> -(2x+1)+2-x>4
=>x<-1 （a)
(ii)-1<x<=2时
原式=》2x+1+2-x>4
=》x>1
=>1<x<=2 （b)
(iii)x>2时
原式=》2x+1+x-2>4
=>x>5/3
=>x>2 (c)
综合(a)(b)(c)=>x<-1或x>1
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x<1
=>0<=x<1或x>2 (a)
(ii)当x<0时
=》x^2+2>-3x
=>x^2+3x+2>0
=>x>-1或x<-2
=>-1<x<0或x<-2 (b)
综合（a)(b)=>-1<x<1或x>2或x<-2

(3)
(x^2+1)(x-3)<=0
因为x^2+1>=1>0
=>x-3<=0
=>x<=3

(4) |3x-4|<=0
因为|3x-4|>=0
=>3x-4=0
=>x=4/3