(创新题)对于函数f(x)定义域中任意x1,x2(x1≠x2)有如下结论:
①f(x1+x2)=f(x1)+f(x2); ②f(x1·x2)=f(x1)+f(x2);
③; ④f()<.
当f(x)=lgx时,上述结论中正确结论的序号是 .
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(创新题)对于函数f(x)定义域中任意x1,x2(x1≠x2)有如下结论:
①f(x1+x2)=f(x1)+f(x2); ②f(x1·x2)=f(x1)+f(x2);
③; ④f()<.
当f(x)=lgx时,上述结论中正确结论的序号是 .