\(A=\frac{1}{2}.\frac{3}{4}.\frac{5}{6}...\frac{79}{80}\)

\(A< \frac{2}{3}.\frac{4}{5}.\frac{6}{7}...\frac{80}{81}\)

\(A^2< \frac{1}{2}.\frac{2}{3}.\frac{3}{4}.\frac{4}{5}.\frac{5}{6}.\frac{6}{7}...\frac{79}{80}.\frac{80}{81}\)

\(A^2< \frac{1}{81}=\left(\frac{1}{9}\right)^2\)

=> \(A< \frac{1}{9}\left(đpcm\right)\)

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< \(\frac{1}{9}\)

\(A=\frac{1}{2}.\frac{3}{4}.\frac{5}{6}...\frac{79}{80}\)

\(A< \frac{2}{3}.\frac{4}{5}.\frac{6}{7}...\frac{80}{81}\)

\(A^2< \frac{1}{2}.\frac{2}{3}.\frac{3}{4}.\frac{4}{5}.\frac{5}{6}.\frac{6}{7}....\frac{79}{80}.\frac{80}{81}\)

\(A^2< \frac{1}{81}=\left(\frac{1}{9}\right)^2\)

\(\Rightarrow A< \frac{1}{9}\left(\text{đ}pcm\right)\)

A = \(\dfrac{1}{2}.\dfrac{3}{4}.\dfrac{5}{6}.....\dfrac{79}{80}\)

=> A₁ < \(\dfrac{2}{3}.\dfrac{4}{5}.\dfrac{5}{6}.....\dfrac{80}{81}\)

=> A² < A.A₁ = \(\dfrac{1}{2}.\dfrac{2}{3}.\dfrac{3}{4}.\dfrac{4}{5}....\dfrac{79}{80}.\dfrac{80}{81}=\dfrac{1}{81}=\left(\dfrac{1}{9}\right)^2\)

=> A < \(\dfrac{1}{9}.\)

Lời giải:

Gọi tổng trên là $A$. Ta có:

\(2A>\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+...+\frac{1}{\sqrt{79}+\sqrt{80}}+\frac{1}{\sqrt{80}+\sqrt{81}}\)

\(2A>\frac{\sqrt{2}-1}{(\sqrt{1}+\sqrt{2})(\sqrt{2}-1)}+\frac{\sqrt{3}-\sqrt{2}}{(\sqrt{2}+\sqrt{3})(\sqrt{3}-\sqrt{2})}+\frac{\sqrt{4}-\sqrt{3}}{(\sqrt{3}+\sqrt{4})(\sqrt{4}-\sqrt{3})}+...+\frac{\sqrt{81}-\sqrt{80}}{(\sqrt{80}+\sqrt{81})(\sqrt{81}-\sqrt{80})}\)

\(2A>(\sqrt{2}-1)+(\sqrt{3}-\sqrt{2})+(\sqrt{4}-\sqrt{3})+....+(\sqrt{81}-\sqrt{80})\)

\(2A>\sqrt{81}-1=8\Rightarrow A>4\)

Ta có đpcm.

\(A=\frac{1}{2}.\frac{3}{4}.\frac{5}{6}....\frac{79}{80}