Đáp án:
$\frac{ 2008 }{2009^{2010} + 1}$ > $\frac{4016}{$2009^{2011}$ -2}$
Giải thích các bước giải:
A = $\frac{2009^{2010} + 2009 }{2009^{2010} + 1}$
= $\frac{2009^{2010} + 1 + 2008 }{2009^{2010} + 1}$
= $\frac{ 2008 }{2009^{2010} + 1}$
B = $\frac{2009^{2010} -2}{2009^{2011} -2}$
= $\frac{2009^{2010} -2 - 4016 }{$2009^{2011}$ -2}$
= $\frac{4016}{$2009^{2011}$ -2}$
Mà $\frac{ 2008 }{2009^{2010} + 1}$ > $\frac{4016}{$2009^{2011}$ -2}$
Nên A > B.
