Ta có : \(M=-\dfrac{7}{10^{2011}}+\dfrac{-15}{10^{2012}}\) và \(N=\dfrac{-15}{10^{2011}}+\dfrac{-8}{10^{2012}}\)
Xét \(M=-\dfrac{7}{10^{2011}}-\dfrac{15}{10^{2012}}=-\dfrac{1}{10^{2011}}\left(7+\dfrac{15}{10}\right)=-\dfrac{1}{10^{2011}}\cdot\dfrac{17}{2}\).
Xét \(N=-\dfrac{15}{10^{2011}}-\dfrac{8}{10^{2012}}=-\dfrac{1}{10^{2011}}\left(15+\dfrac{8}{10}\right)=-\dfrac{1}{10^{2011}}\cdot\dfrac{79}{5}\).
Ta cũng có : \(\dfrac{M}{N}=\dfrac{-\dfrac{1}{10^{2011}}\cdot\dfrac{17}{2}}{-\dfrac{1}{10^{2011}}\cdot\dfrac{79}{5}}=\dfrac{\dfrac{17}{2}}{\dfrac{79}{5}}=\dfrac{85}{158}\)
\(\Rightarrow M=\dfrac{85}{158}N\). Mà \(\dfrac{85}{158}< 1\) nên \(M< N\).
Vậy : \(M< N\).
