题目
题型:不详难度:来源:
x2 |
|m|-1 |
y2 |
m-2 |
答案
x2 |
|m|-1 |
y2 |
m-2 |
∴(|m|-1)(m-2)>0,
解得-1<m<1或m>2,
∴实数m的取值范围是(-1,1)∪(2,+∞).
故答案为:(-1,1)∪(2,+∞).
核心考点
举一反三
x2 |
34 |
y2 |
n2 |
x2 |
n2 |
y2 |
16 |
x2 |
16 |
y2 |
7 |
x2 |
|m|-1 |
y2 |
m-2 |
x2 |
|m|-1 |
y2 |
m-2 |
x2 |
34 |
y2 |
n2 |
x2 |
n2 |
y2 |
16 |
x2 |
16 |
y2 |
7 |