证明：(1)令x=y=0,则f(0)=f(0)f(0)∴f(0)=0或f(0)=1∵f(x)在x∈R上是正的∴f(0)=1令y=-x,则f(x-x)=f(x)f(-x)∴f(-x)=1/f(x),或者f(x)=-f(-x)当x＜0时,f(x)=-f(-x)而-x＞0,即f(-x)∈(0,1)∴f(x)=-f(-x)＞1得证(2)任意...