2A=\(\left(1+\frac{1}{3}\right)\)\(\left(1+\frac{1}{8}\right)\)\(\left(1+\frac{1}{15}\right)\)\(.......\)\(\left(1+\frac{1}{4064255}\right)\)

2A = \(\frac{4}{3}\)\(.\)\(\frac{9}{8}\)\(.\)\(\frac{16}{15}\)\(......\)\(\frac{4064256}{4064255}\)

2A = \(\frac{2.2}{1.3}\)\(.\)\(\frac{3.3}{2.4}\)\(.\)\(\frac{4.4}{3.5}\)\(......\)\(\frac{2016.2016}{2015.2017}\)

2A = \(\frac{2.3.4....2016}{1.2.3.....2015}\)\(.\)\(\frac{2.3.4....2016}{3.4.5....2017}\)

2A = \(\frac{2016}{1}\)\(.\)\(\frac{2}{2017}\)

2A = \(\frac{4032}{2017}\)

A = \(\frac{4032}{2017}\)\(:2\)

A = \(\frac{2016}{2017}\)

\(\frac{1}{2}.\left(1+\frac{1}{1.3}\right).\left(1+\frac{1}{2.4}\right).\left(1+\frac{1}{3.5}\right)...\left(1+\frac{1}{2015.2017}\right)\)

\(=\frac{1}{2}.\frac{1.3+1}{1.3}.\frac{2.4+1}{2.4}.\frac{3.5+1}{3.5}...\frac{2015.2017+1}{2015.2017}\)

\(=\frac{1}{2}.\frac{2.2}{1.3}.\frac{3.3}{2.4}.\frac{4.4}{3.5}...\frac{2016.2016}{2015.2017}\)

\(=\frac{1}{2}.\frac{2.3.4...2016}{1.2.3...2015}.\frac{2.3.4...2016}{3.4.5...2017}\)

\(=\frac{1}{2}.2016.\frac{2}{2017}=\frac{2016}{2017}\)

\(\frac{1}{2}.\left(1+\frac{1}{1.3}\right).\left(1+\frac{1}{2.4}\right).\left(1+\frac{1}{3.5}\right)...\left(1+\frac{1}{2015.2017}\right)\)

\(=\frac{1}{2}.\frac{1.3+1}{1.3}.\frac{2.4+1}{2.4}.\frac{3.5+1}{3.5}...\frac{2015.2017+1}{2015.2017}\)

\(=\frac{1}{2}.\frac{2.2}{1.3}.\frac{3.3}{2.4}.\frac{4.4}{3.5}...\frac{2016.2016}{2015.2017}\)

\(=\frac{1}{2}.\frac{2.3.4...2016}{1.2.3...2015}.\frac{2.3.4...2016}{3.4.5...2017}\)

\(=\frac{1}{2}.2016.\frac{2}{2017}=\frac{2016}{2017}\)

a: \(\Leftrightarrow\dfrac{x-214}{86}-1+\dfrac{x-132}{84}-2+\dfrac{x-54}{82}-3=0\)

=>x-300=0

hay x=300

\(\left(a+\frac{1}{1.3}\right)+\left(a+\frac{1}{3.5}\right)+...+\left(a+\frac{1}{23.25}\right)=11a+\left(\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}\right)\)

\(\Rightarrow12a+\left(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{23.25}\right)=11a+\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+\frac{1}{3^5}\right)\)(1)

Ta có \(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{23.25}=\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{23.25}\right)\)

\(=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{23}-\frac{1}{25}\right)=\frac{1}{2}\left(1-\frac{1}{25}\right)=\frac{1}{2}.\frac{24}{25}=\frac{12}{25}\)

Lại có \(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+\frac{1}{3^5}=\frac{3\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+\frac{1}{3^5}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+\frac{1}{3^5}\right)}{2}\)

\(=\frac{1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}-\frac{1}{3}-\frac{1}{3^2}-\frac{1}{3^3}-\frac{1}{3^4}-\frac{1}{3^5}}{2}=\frac{1-\frac{1}{3^5}}{2}=\frac{1}{2}-\frac{1}{3^5.2}\)

Khi đó (1) <=> \(12a-\frac{12}{25}=11a+\frac{1}{2}-\frac{1}{3^5.2}\)

=> \(a=\frac{12}{25}+\frac{1}{2}-\frac{1}{3^5.2}=\frac{49}{50}-\frac{1}{3^5.2}=\frac{49}{50}-\frac{1}{486}=\frac{23764}{24300}\)

Gọi \(A=\left(a+\frac{1}{1.3}\right)+\left(a+\frac{1}{3.5}\right)+\left(a+\frac{1}{5.7}\right)+...+\left(a+\frac{1}{23.25}\right)\)

\(\Rightarrow A=12a+\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{23.25}\right)\)

\(\Rightarrow A=12a+\left[\frac{1}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{23.25}\right)\right]\)

\(\Rightarrow A=12a+\left[\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{23}-\frac{1}{25}\right)\right]\)

\(\Rightarrow A=12a+\left[\frac{1}{2}\left(1-\frac{1}{25}\right)\right]\)

\(\Rightarrow A=12a+\left(\frac{1}{2}.\frac{24}{25}\right)\)

\(\Rightarrow A=12a+\frac{12}{25}\)

Gọi \(B=\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}\)

\(\Rightarrow B=\frac{1}{1.3}+\frac{1}{3.3}+\frac{1}{9.3}+\frac{1}{27.3}+\frac{1}{81.3}\)

\(\Rightarrow3B=1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}\)

\(\Rightarrow3B-B=1-\frac{1}{243}\)

\(\Rightarrow2B=\frac{242}{243}\)

\(\Rightarrow B=\frac{121}{243}\)

\(\Rightarrow A=11a+B\)

\(\Rightarrow12a+\frac{12}{25}=11a+\frac{121}{243}\)

\(\Leftrightarrow12a-11a=\frac{121}{243}-\frac{12}{25}\)

\(\Leftrightarrow a=\frac{109}{6075}\)