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由於|x-1|+|x-2|表示數軸上的x對應點到1、2對應點的距離之和,而0和3對應點到1、2對應點的距離之和等於3,故當x<1,或x>3時,不等式|x-1|+|x-2|>3成立.故不等式|x-1|+|x-2|>3的解集為(-∞,0)∪(3,+∞),故答案為(-∞,0)∪(3,+∞).
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