2a^2+a+1=2(a^2+a/2)+1=2(a+1/4)^2+7/8 恒大于0
-3a^2+2a-1=-3(a^2-2a/3)-1=-3(a-1/3)^2-2/3 恒小于0
所以-(-3a^2+2a-1)=3a^2-2a+1 恒大于0
因为为偶函数,有f(x)=f(-x)
所以f(-3a^2+2a-1)=f(3a^2-2a+1)
所以有f(2a^2+a+1)3a^2-2a+1
a^2-3a
已知f(x)是定义在R上的偶函数,在区间(0,+∞)上是减函数,且f(2a^2+a+1)
已知f(x)是定义在R上的偶函数,在区间(0,+∞)上是减函数,且f(2a^2+a+1)
数学人气:988 ℃时间:2019-10-20 00:12:38
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