少个条件吧：当x1.非零函数f(x)对实数a,b均有f(a+b)=f(a)*f(b)且当x1.当f(4)=1/16时,解不等式f(x-3)*f(5) f(x)=f(x/2)^2>0(因为函数是非零函数,f(x/2)≠0)2.令a=0 b=0f(0)=f(0)^2 ,f(0)=1令a=x ,b=-x ,f(0)=f(x)f(-x...