Question

cho a =1/2*3/4*5/6*...*79/80. chứng minh a <1/9

Answer 1

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

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

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

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

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

Answer 2

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< \(\frac{1}{9}\)

Answer 3

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

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

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

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

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