Question

對于0≤m≤4的m，不等式x²+mx＞4x+m-3恒成立，則x的取值范圍是______．

Answer 1

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

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．
故答案為：x＞3或x＜-1