已知p:(x-3)的绝对值

p:|x-3|≤2-2≤x-3≤21≤x≤5所以非p:x<1或者x>5q:(x-m+1)(x-m-1)≤0(x-m)² - 1 ≤ 0(x-m)²≤ 1-1 ≤ x-m≤ 1m-1 ≤ x ≤ m+1所以非q:x<m-1 或者x>m+1非p是非q的充分而不必要条件,即非p能推出...