6．已知f(x+1)f(x+1)f(x+1)是定义在RRR上的偶函数，且对任意的1⩽x1<x21\leqslant x_{1}<x_{2}1⩽x1<x2，都有(x1−x2)[f(x1)−f(x2)]<0(x_{1}-x_{2})[f(x_{1})-f(x_{2})]<0(x1−x2)[f(x1)−f(x2)]<0恒成立，则关于xxx的不等式f(2x)>f(x−1)f(2x)>f(x-1)f(2x)>f(x−1)的解集为((( )))