$(log_x{8}+ log_4{x^2}) log_2{\sqrt{2x}} \geq 0$
Giải phương trình $(log_x{8}+ log_4{x^2}) log_2{\sqrt{2x}} \geq 0$
Started By Linhchuachua, 15-03-2015 - 23:17
#1
Posted 15-03-2015 - 23:17
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