(1)证明:设x1,x2∈(0,1)且x1<x 2 则,
f(x1) f(x2)=2x14x1+12x24x2+1=2x1(4x2+1) 2x2(4x1+1)(4x1+1)(4x2+1)=(2x2 2x1)(2x1+x21) (4x1+1)(4x2+1)…(3分)
∵0<x1<x 2<1,∴2x2>2x1 ,2x1+x2>1
∴f(x1)-f(x2)>0...
(1)证明:设x1,x2∈(0,1)且x1<x 2 则,
f(x1) f(x2)=2x14x1+12x24x2+1=2x1(4x2+1) 2x2(4x1+1)(4x1+1)(4x2+1)=(2x2 2x1)(2x1+x21) (4x1+1)(4x2+1)…(3分)
∵0<x1<x 2<1,∴2x2>2x1 ,2x1+x2>1
∴f(x1)-f(x2)>0...