设函数f(x)在R上是偶函数,在区间（-∞,0）上递增,且f(2a²+a+1）＜f(2a²-2a+3

2a²+a+1=2(a+1/4)^2+7/8>0
2a²-2a+3=2(a-1/2)^2+5/2>0
函数f(x)在R上是偶函数,在区间（-∞,0）上递增,所以在（0,∞）上递减
2a²+a+1>2a²-2a+3
a>2/3