请问这个不等式怎么解呀

用b 表示x 的范围
2025-01-07 21:55:52
推荐回答(3个)
回答1:

跟你说个思路吧,两边同时乘以ex,然后把右边的(2b-2)ex移项到左边,把ex用t代替,记住t是大于0的。然后就是解关于t的二次不等式:t2-2(b-1)t+1<0
这时候分情况讨论了:
一种是上式无解,既最小值(t=b-1对应的值)大于或等于0,解得0≤b≤2;
即0≤b≤2时,x∈∞
另一种是上式有解,
b>2时,t=b-1>0,对称轴在右边,二次不等式对应的两个解就是答案,即b-√ ̄(b2-2b)-1<t<b+√ ̄(b2-2b)-1,然后把ex带进去解出来就好了;
b<0时,t=b-1<0,对称轴在左边,因为t>0,所以t无解,即x无解。
解答这道题还是要通过图形来直观反映,多画图。

回答2:

换元 用对勾函数性质求解

回答3:

提示一下。
首先,分离变量。
然后利用基本不等式。

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