|x+2|+|x+3|≥5
case 1: x<-3
|x+2|+|x+3|≥5
-(x+2)-(x+3)≥5
-2x-5≥5
x≤-5
solution for case 1: x≤-5
case 2: x=-3
|x+2|+|x+3|≥5 : false
case 3:-3< x<2
|x+2|+|x+3|≥5
-(x+2)+(x+3)≥5
1≥5
false
case 4: x=-2
|x+2|+|x+3|≥5 : false
case 5: x>-2
|x+2|+|x+3|≥5
(x+2)+(x+3)≥5
x≥0
solution for case 5: x≥0
|x+2|+|x+3|≥5
case 1 or case 2 or case 3 or case 4 or case 5
x≤-5 or x≥0
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