函数f(x)=alog2^x-blog3^x+3,令g(x)=alog2^x-blog3^x,f(x)=g(x)+3f(2010)=g(2010)+3=4g(2010)=1g(1/x)=alog2^(1/x)-blog3^(1/x)=alog2^x^(-1)-blog3^x^(-1)= -[alog2^x-blog3^x]=-g(x)g(1/2010)=-g(2010)=-1,则 f(1/...