\(A=\frac{1}{2}.\frac{3}{4}.\frac{5}{6}...\frac{79}{80}\)
\(A< \frac{2}{3}.\frac{4}{5}.\frac{6}{7}...\frac{80}{81}\)
\(A^2< \frac{1}{2}.\frac{2}{3}.\frac{3}{4}.\frac{4}{5}.\frac{5}{6}.\frac{6}{7}...\frac{79}{80}.\frac{80}{81}\)
\(A^2< \frac{1}{81}=\left(\frac{1}{9}\right)^2\)
=> \(A< \frac{1}{9}\left(đpcm\right)\)
Ta có:
\(\frac{1}{2}\)= 1- \(\frac{1}{2}\) < 1- \(\frac{1}{3}\)=\(\frac{2}{3}\)
\(\frac{3}{4}\)= 1- \(\frac{1}{4}\) < 1- \(\frac{1}{5}\) = \(\frac{4}{5}\)
...
\(\frac{79}{80}\) = 1- \(\frac{1}{80}\) < 1- \(\frac{1}{81}\)= \(\frac{80}{81}\)
Từ trên, ta có:
A= \(\frac{1}{2}\). \(\frac{3}{4}\). \(\frac{5}{6}\)...\(\frac{79}{80}\)< \(\frac{2}{3}\). \(\frac{4}{5}\). \(\frac{6}{7}\)...\(\frac{80}{81}\)
A2 < \(\left(\frac{2}{3}.\frac{4}{5}.\frac{6}{7}...\frac{80}{81}\right)\). \(\left(\frac{1}{2}.\frac{3}{4}.\frac{5}{6}...\frac{79}{80}\right)\)
A2 < \(\frac{1}{2}.\frac{2}{3}.\frac{3}{4}...\frac{79}{80}.\frac{80}{81}\)
A2 <\(\frac{1.\left(2.3.4...79.80\right)}{\left(2.3.4...79.80\right).81}\)
A2 < \(\frac{1}{81}\) =\(\left(\frac{1}{9}\right)^2\)
A < \(\frac{1}{9}\) (đpcm)
Vậy A< \(\frac{1}{9}\)
\(A=\frac{1}{2}.\frac{3}{4}.\frac{5}{6}...\frac{79}{80}\)
\(A< \frac{2}{3}.\frac{4}{5}.\frac{6}{7}...\frac{80}{81}\)
\(A^2< \frac{1}{2}.\frac{2}{3}.\frac{3}{4}.\frac{4}{5}.\frac{5}{6}.\frac{6}{7}....\frac{79}{80}.\frac{80}{81}\)
\(A^2< \frac{1}{81}=\left(\frac{1}{9}\right)^2\)
\(\Rightarrow A< \frac{1}{9}\left(\text{đ}pcm\right)\)
