若不等式x²+mx＞4x+m-3恒成立
则m（x-1）+x²-4x+3＞0在0≤m≤4时恒成立．
令f（m）=m（x-1）+x²-4x+3．
则

f(0)＞0 f(4)＞0

⇒

x2−4x+3＞0 x2−1＞0

⇒

x＜1或x＞3 x＜−1或x＞1.

∴x＜-1或x＞3．
故答案为：x＞3或x＜-1