Đáp án: $\dfrac{7^{2008} + 1}{7^{2009} +1} > \dfrac{7^{2009} + 1}{7^{2010} +1}$
Giải thích các bước giải:
Ta có:
$\dfrac{7^{2008} + 1}{7^{2009} +1}$
$= \dfrac{1}{7}. \dfrac{7^{2009} + 7}{7^{2009} +1}$
$= \dfrac{1}{7} . \dfrac{7^{2009} + 1 + 6}{7^{2009} + 1}$
$= \dfrac{1}{7} (1 + \dfrac{6}{7^{2009} +1})$
$\dfrac{7^{2009} + 1}{7^{2010} +1}$
$= \dfrac{1}{7}. \dfrac{7^{2010} + 7}{7^{2010} +1}$
$= \dfrac{1}{7} . \dfrac{7^{2010} + 1 + 6}{7^{2010} + 1}$
$= \dfrac{1}{7} (1 + \dfrac{6}{7^{2010} +1})$
Vì : $ \dfrac{6}{7^{2009} +1} > \dfrac{6}{7^{2010} +1}$
$⇒ \dfrac{7^{2008} + 1}{7^{2009} +1} > \dfrac{7^{2009} + 1}{7^{2010} +1}$
