∵|x-2|＜x2，∴-x2＜x-2＜x2，∴ x2+x?2＞0① x2?x+2＞0② ，解①得x＜-2或x＞1；解②得x∈R；∴绝对值不等式|x-2|＜x2的解集是（-∞，-2）∪（1，+∞）．故答案为：（-∞，-2）∪（1，+∞）．
