1.设x2>x1>0则x2/x1>1 ,f(x2/x1)>0f(x2)=f(x1*x2/x1)=f(x1)+f(x2/x1)f(x2)-f(x1)>0得证2.if x1=x2=1 then f(1*1)=f(1)+f(1)=>f(1)=0if x1=x2=2 then f(2*2)=f(2)+f(2)=>f(4)=2f(2x^2-1)解:0<2x^2-1<4=>x^2∈(1/2,5/2)
