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a=\(\dfrac{1}{2}\cdot\dfrac{3}{4}\cdot\dfrac{5}{6}\cdot\dfrac{7}{8}\cdot...\cdot\dfrac{79}{80}\)
a<\(\dfrac{2}{3}\cdot\dfrac{4}{5}\cdot\dfrac{6}{7}\cdot\dfrac{8}{9}\cdot...\cdot\dfrac{80}{81}\)
\(\text{a}^2< \dfrac{1}{2}\cdot\dfrac{2}{3}\cdot\dfrac{3}{4}\cdot\dfrac{4}{5}\cdot\dfrac{5}{6}\cdot\dfrac{6}{7}\cdot\dfrac{7}{8}\cdot\dfrac{8}{9}\cdot...\cdot\dfrac{79}{80}\cdot\dfrac{80}{81}\)
\(\Rightarrow\text{a}^2< \dfrac{1}{81}=\left(\dfrac{1}{9}\right)^2\)
\(\Rightarrow\text{a}< \dfrac{1}{9}\)(dpcm)
Nho tich cho mk nhe
Ta đặt B=\(\frac{2}{3}.\frac{4}{5}...\frac{80}{81}\)
Mà \(\frac{1}{2}<\frac{2}{3};\frac{3}{4}<\frac{4}{5};...;\frac{79}{80}<\frac{80}{81}\)
=>A<B
=>A2<AB=\(\frac{1}{2}.\frac{2}{3}.....\frac{80}{81}=\frac{1}{81}\)
=>A2<\(\frac{1}{81}\)
=>A<\(\sqrt{\frac{1}{81}}=\frac{1}{9}\)(đpcm)
\(A=\frac{1}{2}.\frac{3}{4}.\frac{5}{6}...\frac{79}{80}\)
\(A< \frac{2}{3}.\frac{4}{5}.\frac{6}{7}...\frac{80}{81}\)
\(A^2< \frac{1}{2}.\frac{2}{3}.\frac{3}{4}.\frac{4}{5}.\frac{5}{6}.\frac{6}{7}...\frac{79}{80}.\frac{80}{81}\)
\(A^2< \frac{1}{81}=\left(\frac{1}{9}\right)^2\)
=> \(A< \frac{1}{9}\left(đpcm\right)\)
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< \(\frac{1}{9}\)
ta có 1/2 * 3/ 4 * 5/6 *... * 79/80 = 0.0889
so sánh a với 1/9
0.0889 < 0.(1)
=> A < 1/9
A = 1/80
so sánh a với 1/9
1/80 <1/9