（1）y=x+2+1x2−x−6的定义域满足：x+2≥0x2−x−6≠0，解得x＞-2且x≠3．∴y=x+2+1x2−x−6的定义域为（-2，3）∪（3，+∞）．（2）y=(x+1)0|x|−x的定义域满足：x+1≠0|x|−x≠0，解得x＜0且x≠-1，∴y=(x+1)0|x|...