若(log23)x-(log53)x≥(log23)-y-(log53)-y,则
A.x-y≥0
B.x+y≥0
C.x-y≤0
D.x+y≤0
网友回答
B解析分析:令F(x)=(log23)x-(log53)x,然后根据复合函数的单调性法则确定F(x)的单调性,最后根据单调性解F(x)≥F(-y)即可.解答:令F(x)=(log23)x-(log53)x∵log23>1,0<log53<1∴函数F(x)在R上单调递增∵(log23)x-(log53)x≥(log23)-y-(log53)-y,∴F(x)≥F(-y)∴x≥-y即x+y≥0故选B.点评:本题主要考查了对数函数的单调性,以及复合函数的单调性和构造法的运用,属于中档题.