方法1：展开法当x ≤1时,|x-1|+|x-5| = 1-x+5-x = 6-2x,最小值在x=1处取得,最小值为4当1＜x＜5时,|x-1|+|x-5| = x - 1+5-x = 4,最小值为4当x≥5时,|x-1|+|x-5| = x-1+x-5 = 2x - 6,最小值在x=5时取得,最小值为4所以|...