Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
a,\\
\frac{{n - 1}}{n} - \frac{n}{{n + 1}} = \frac{{\left( {n - 1} \right).\left( {n + 1} \right) - {n^2}}}{{n\left( {n + 1} \right)}} = \frac{{{n^2} - 1 - {n^2}}}{{n\left( {n + 1} \right)}} = \frac{{ - 1}}{{n\left( {n + 1} \right)}} < 0\\
\Rightarrow \frac{{n - 1}}{n} < \frac{n}{{n + 1}}\\
\Rightarrow \left\{ \begin{array}{l}
\frac{1}{2} < \frac{2}{3}\\
\frac{3}{4} < \frac{4}{5}\\
\frac{5}{6} < \frac{6}{7}\\
....\\
\frac{{99}}{{100}} < \frac{{100}}{{101}}
\end{array} \right.\\
\Rightarrow \frac{1}{2}.\frac{3}{4}.\frac{5}{6}.....\frac{{99}}{{100}} < \frac{2}{3}.\frac{4}{5}.\frac{6}{7}.....\frac{{100}}{{101}}\\
\Leftrightarrow M < N\\
b,\\
M.N = \left( {\frac{1}{2}.\frac{3}{4}.\frac{5}{6}.....\frac{{99}}{{100}}} \right).\left( {\frac{2}{3}.\frac{4}{5}.\frac{6}{7}.....\frac{{100}}{{101}}} \right)\\
= \frac{1}{2}.\frac{2}{3}.\frac{3}{4}.\frac{4}{5}.....\frac{{99}}{{100}}.\frac{{100}}{{101}}\\
= \frac{1}{{101}}\\
c,\\
M.N = \frac{1}{{101}} < \frac{1}{{100}}\\
M < N \Rightarrow \frac{1}{{100}} > M.N > M.M = {M^2}\\
\Rightarrow \frac{1}{{10}} > M \Leftrightarrow M < \frac{1}{{10}}
\end{array}\)
