Question

高中数学求过程

Answer 1

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: 闹槐盯 x≥1

|x-1|-|2x+1|+3≤0

case 1 or case 2 or case 3

 x≤-5 or 液和x≥1

ie

2^(|x-1|)≤(1/2)^(3-|2x+1|) 

 => x≤明旁-5 or x≥1