3．定义在RRR上的偶函数f(x)f(x)f(x)满足：对任意的x1x_{1}x1，x2∈(−∞x_{2}\in (-\inftyx2∈(−∞，0](x1≠x2)0](x_{1}\ne x_{2})0](x1=x2)，有f(x1)−f(x2)x1−x2<0\frac{f({x_1})-f({x_2})}{{x_1}-{x_2}}<0x1−x2f(x1)−f(x2)<0，则((( )))