如何推出<0
(3n+3)/(2n²+5n+2)-(3n+4)/(2n²+5n+3)<(3n+3)/(2n²+5n+3)-(3n+4)/(2n²+5n+3)=-1/(2n²+5n+3)由上式可知分母恒为正值,分子为负一,所以原式<0