高中数学求过程
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2^(|x-1|)≤(1/2)^(3-|2x+1|)
2^(|x-1|)≤2^(|2x+1|-3)
|x-1|≤|2x+1|-3
|x-1|-|2x+1|+3≤0
case 1: x≤-1/2
|x-1|-|2x+1|+3≤0
-(x-1)+(2x+1)+3≤0
x+5≤0
x≤-5
solution for case 1: x≤-5
case 2: -1/2<x<1
|x-1|-|2x+1|+3≤0
-(x-1)-(2x+1)+3≤0
-3x+3≤0
x≥1
solution for case 2: no solution
case 3: x≥1
|x-1|-|2x+1|+3≤0
(x-1)-(2x+1)+3≤0
-x+1≤0
x≥1
solution for case 3: x≥1
|x-1|-|2x+1|+3≤0
case 1 or case 2 or case 3
x≤-5 or x≥1
ie
2^(|x-1|)≤(1/2)^(3-|2x+1|)
=> x≤-5 or x≥1
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