②f(x1•x2)=lgx1x2;f(x1)+f(x2)=lgx1+lgx2=lgx1x2;故正确;
③∵f(x)=lgx在定义域上单调递增,则当x1<x2时,f(x1)<f(x2);则
f(x1)-f(x2) |
x1-x2 |
④∵f(
x1+x2 |
2 |
x1+x2 |
2 |
f(x1)+f(x2) |
2 |
lgx1+lgx2 |
2 |
x1x2 |
又∵
x1+x2 |
2 |
x1x2 |
则lg
x1+x2 |
2 |
x1x2 |
故选②③.
f(x1)-f(x2) |
x1-x2 |
x1+x2 |
2 |
f(x1)+f(x2) |
2 |
f(x1)-f(x2) |
x1-x2 |
x1+x2 |
2 |
x1+x2 |
2 |
f(x1)+f(x2) |
2 |
lgx1+lgx2 |
2 |
x1x2 |
x1+x2 |
2 |
x1x2 |
x1+x2 |
2 |
x1x2 |