Cho biết \(M = \dfrac{{{7^{2013}} + 1}}{{{7^{2014}} + 1}}\) và  \(N = \dfrac{{{7^{2014}} + 1}}{{{7^{2015}} + 1}}\). So sánh \(M\) và \(N\) khẳng định nào sau đây là khẳng định đúng.

Ta có:

\(\begin{array}{l}7M = \dfrac{{7.\left( {{7^{2013}} + 1} \right)}}{{{7^{2014}} + 1}} = \dfrac{{{{7.7}^{2013}} + 7.1}}{{{7^{2014}} + 1}}\\ \,\,\,\,\,\,\,\,\,= \dfrac{{{7^{2014}} + 7}}{{{7^{2014}} + 1}} = \dfrac{{{7^{2014}} + 1 + 6}}{{{7^{2014}} + 1}} \\\,\,\,\,\,\,\,\,\,= \dfrac{{{7^{2014}} + 1}}{{{7^{2014}} + 1}} + \dfrac{6}{{{7^{2014}} + 1}} \\\,\,\,\,\,\,\,\,\,= 1 + \dfrac{6}{{{7^{2014}} + 1}}\end{array}\)

\(\begin{array}{l}7N = \dfrac{{7.\left( {{7^{2014}} + 1} \right)}}{{{7^{2015}} + 1}} = \dfrac{{{{7.7}^{2014}} + 7.1}}{{{7^{2015}} + 1}}\\\,\,\,\,\,\,\,\,\, = \dfrac{{{7^{2015}} + 7}}{{{7^{2015}} + 1}}= \dfrac{{{7^{2015}} + 1 + 6}}{{{7^{2015}} + 1}}\\\,\,\,\,\,\,\,\,\, = \dfrac{{{7^{2015}} + 1}}{{{7^{2015}} + 1}} + \dfrac{6}{{{7^{2015}} + 1}}\\\,\,\,\,\,\,\,\,\, = 1 + \dfrac{6}{{{7^{2015}} + 1}}\end{array}\)

Vì \({7^{2015}} + 1 > {7^{2014}} + 1\)  nên \(\dfrac{6}{{{7^{2014}} + 1}} > \dfrac{6}{{{7^{2015}} + 1}}\)

Do đó \(7M > 7N\) hay \(M > N.\)