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in Linear Programming by (20 points)
  1. solve 1≤|x-2|≤3

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1 Answer

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by (24.8k points)

x−2<0
x<2
1≤−x+2≤3
1−2≤−x+2−2≤3−2
−1≤−x≤1
1≥x≥−1
x∈[−1,1]
x−2>0
x>2
1≤x−2≤3
3≤x≤5
x∈[3,5]
x∈[−1,1]U[3,5]

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