∵对任意的x∈R都有f(x+2)=f(x),∴函数是以2为周期的周期函数; 根据若0≤x 1 <x 2 ≤1,都有f(x 1 )>f(x 2 ),函数在区间[0,1]上是减函数; ∵y=f(x+1)是偶函数,∴f(-x+1)=f(x+1),其图象关于x=1直线对称, ∴f(-2)=f(0); f(7.8)=f(6+1.8)=f(1.8)=f(0.8+1)=f(-0.8+1)=f(0.2); f(5.5)=f(4+1.5)=f(1.5)=f(0.5+1)=f(-0.5+1)=f(0.5); ∵0≤x 1 <x 2 ≤1,都有f(x 1 )>f(x 2 ); ∴f(-2)>f(7.8)>f(5.5). 故选B |