Ta có: a = (1 - 1/2) + (1 - 1/4) + (1 - 1/6) +...+ (1 - 1/80)

             =  (1 + 1 + 1 +...+ 1) - (1/2 + 1/4 + 1/6 + ... + 1/80)

             = 40 - ... 

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 < \(\dfrac{1}{9}.\)

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

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

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

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

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

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=\(\dfrac{1}{2}\cdot\dfrac{3}{4}\cdot\dfrac{5}{6}\cdot\dfrac{7}{8}\cdot...\cdot\dfrac{79}{80}\)

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

a)\(-\dfrac{2}{5}.\dfrac{4}{7}+\dfrac{-3}{5}.\dfrac{2}{7}+\dfrac{-3}{5}\) 

=\(-\dfrac{2}{5}.\dfrac{4}{7}+\dfrac{3}{7}.\dfrac{-2}{5}+\dfrac{-3}{5}\)

=\(-\dfrac{2}{5}.\left(\dfrac{4}{7}+\dfrac{3}{7}\right)+\dfrac{-3}{5}\) 

=\(\dfrac{-2}{5}.1+\dfrac{-3}{5}\) 

=\(-\dfrac{2}{5}+\dfrac{-3}{5}\) 

=\(-\dfrac{5}{5}\) = -1

\(\dfrac{5}{9}.\dfrac{14}{17}+\dfrac{1}{17}.\dfrac{5}{9}+\dfrac{2}{9}+\dfrac{5}{12}\)

=\(\dfrac{5}{9}.\left(\dfrac{14}{17}+\dfrac{1}{17}\right)+\dfrac{2}{9}+\dfrac{5}{12}\) 

=\(\dfrac{5}{9}.\dfrac{15}{17}+\dfrac{2}{9}+\dfrac{5}{12}\) 

=\(\dfrac{25}{51}+\dfrac{2}{9}+\dfrac{5}{12}\)

=\(\dfrac{691}{612}\)

\(-\dfrac{4}{25}:\dfrac{11}{79}+\dfrac{15}{11}:\dfrac{25}{79}\)

=\(-\dfrac{4}{25}.\dfrac{79}{11}+\dfrac{15}{11}.\dfrac{79}{25}\)

=\(-\dfrac{4}{25}.\dfrac{79}{11}+\dfrac{15}{25}.\dfrac{79}{11}\) 

=\(\left(-\dfrac{4}{25}+\dfrac{15}{25}\right).\dfrac{79}{11}\)

=\(\dfrac{11}{25}.\dfrac{79}{11}\) 

=\(\dfrac{79}{25}\)

\(6+6^2+\cdot\cdot\cdot+6^{10}\)

\(=6\cdot\left(1+6\right)+6^3\cdot\left(1+6\right)+\cdot\cdot\cdot+6^9\cdot\left(1+6\right)\)

\(=6\cdot7+6^3\cdot7+\cdot\cdot\cdot+6^9\cdot7\)

\(=7\cdot\left(6+6^3+\cdot\cdot\cdot+6^9\right)⋮7\)

\(\Rightarrow6+6^2+\cdot\cdot\cdot\cdot+6^{10}⋮7\)

\(5^1-5^9+5^8=5\left(1-5^8+5^7\right)⋮7\Leftrightarrow5^8-5^7-1⋮7\)

\(5\equiv-2\left(mod7\right)\Rightarrow5^3\equiv-1\left(mod7\right)\Rightarrow5^8\equiv4\left(mod7\right);5^7\equiv-2\left(mod7\right)\)

\(5^8-5^7-1\equiv5\left(mod7\right):v\)

\(6+6^2+\cdot\cdot\cdot+6^{10}\)

\(=6\cdot7+6^3\cdot7+\cdot\cdot\cdot+6^9\cdot7\)

\(=7\cdot\left(6+\cdot\cdot\cdot+6^9\right)\)

\(⋮7\)