CMR: $\frac{1}{3}$ . $\frac{4}{6}$ . $\frac{7}{9}$…$\frac{100}{102}$ < $\frac{1}{17}$
CMR: $\frac{1}{3}$ . $\frac{4}{6}$ . $\frac{7}{9}$…$\frac{100}{102}$ < $\frac{1}{17}$
By Vivian
By Vivian
CMR: $\frac{1}{3}$ . $\frac{4}{6}$ . $\frac{7}{9}$…$\frac{100}{102}$ < $\frac{1}{17}$
Giải thích các bước giải:
\(\begin{array}{l}
A = \frac{1}{3}.\frac{4}{6}.\frac{7}{9}…\frac{{100}}{{102}}\\
CM:\frac{n}{{n + 2}} < \frac{{n – 1}}{n}(n \ge 4)\\
= > \frac{4}{6} < \frac{3}{4}\\
\frac{7}{9} < \frac{6}{7}\\
….\\
\frac{{100}}{{102}} < \frac{{99}}{{100}}\\
= > {A^2} < \frac{1}{{{3^2}}}.\frac{3}{4}.\frac{4}{6}.\frac{6}{7}…\frac{{99}}{{100}}.\frac{{100}}{{102}}\\
= > {A^2} < \frac{1}{{918}}\\
= > A < \sqrt {\frac{1}{{918}}} < \frac{1}{{17}}\\
\end{array}\)