\(A=\dfrac{1}{2}\left(2.\dfrac{2}{3}\right)\left(\dfrac{3}{2}.\dfrac{3}{4}\right)\left(\dfrac{4}{3}.\dfrac{4}{5}\right)....\left(\dfrac{2016}{2015}.\dfrac{2016}{2017}\right)\)

\(=\dfrac{2016}{2017}\)

\(A=\dfrac{1}{2}.\left(1+\dfrac{1}{1.3}\right)\left(1+\dfrac{1}{2.4}\right)\left(1+\dfrac{1}{3.5}\right)....\left(\dfrac{1}{2015.2017}\right)\)

\(=\dfrac{1}{2}\left(\dfrac{2}{1}.\dfrac{2}{3}\right).\left(\dfrac{3}{2}.\dfrac{3}{4}\right).\left(\dfrac{4}{3}.\dfrac{4}{5}\right)....\left(\dfrac{2016}{2015}.\dfrac{2016}{2017}\right)\)

\(=\dfrac{1}{2}.\left(\dfrac{2}{1}.\dfrac{2}{3}\right).\left(\dfrac{3}{2}.\dfrac{3}{4}\right).\left(\dfrac{4}{3}.\dfrac{4}{5}\right).....\left(\dfrac{2016}{2015}.\dfrac{2016}{2017}\right)\)

\(=\dfrac{2016}{2017}\)

\(A=\dfrac{1}{2}\left(\dfrac{2.2}{1.3}\right).\left(\dfrac{3.3}{2.4}\right)...\left(\dfrac{2020.2020}{2019.2021}\right)\)

\(=\dfrac{1.2.2.3.3...2020.2020}{1.2.2.3.3.4.4...2019.2021}\)

\(=\dfrac{1}{2021}\)

\(A=\dfrac{1}{2}\cdot\left(1+\dfrac{1}{1\cdot3}\right)\left(1+\dfrac{1}{2\cdot4}\right)\left(1+\dfrac{1}{3\cdot5}\right)...\left(1+\dfrac{1}{2019\cdot2021}\right)\)

\(A=\dfrac{1}{2}\left(1+\dfrac{1}{2^2-1}\right)\left(1+\dfrac{1}{3^2-1}\right)\left(1+\dfrac{1}{4^2-1}\right)...\left(1+\dfrac{1}{2020^2-1}\right)\)

\(A=\dfrac{1}{2}\cdot\dfrac{2^2}{\left(2-1\right)\left(2+1\right)}\cdot\dfrac{3^2}{\left(3-1\right)\cdot\left(3+1\right)}...\left(\dfrac{2020^2}{\left(2020-1\right)\cdot\left(2020+1\right)}\right)\)

\(A=\dfrac{1}{2}\cdot\dfrac{2}{1}\cdot\dfrac{2}{3}\cdot\dfrac{3}{2}\cdot\dfrac{3}{4}\cdot...\cdot\dfrac{2020}{2019}\cdot\dfrac{2020}{2021}\)

\(A=\dfrac{1}{2}\cdot\dfrac{2}{1}\cdot\dfrac{3}{2}\cdot...\cdot\dfrac{2020}{2019}\cdot\dfrac{2}{3}\cdot\dfrac{3}{4}\cdot...\cdot\dfrac{2020}{2021}\)

\(A=\dfrac{1}{2}\cdot2020\cdot\dfrac{2}{2021}=\dfrac{2020}{2021}\)

Sửa đề: A=(1+1/1*3)(1+1/2*4)*...*(1+1/2019*2021)

\(=\dfrac{2^2}{\left(2-1\right)\left(2+1\right)}\cdot\dfrac{3^2}{\left(3-1\right)\left(3+1\right)}\cdot...\cdot\dfrac{2020^2}{\left(2020-1\right)\left(2020+1\right)}\)

\(=\dfrac{2}{1}\cdot\dfrac{3}{2}\cdot...\cdot\dfrac{2020}{2019}\cdot\dfrac{2}{3}\cdot\dfrac{3}{4}\cdot...\cdot\dfrac{2020}{2021}=2020\cdot\dfrac{2}{2021}=\dfrac{4040}{2021}\)

\(=\dfrac{1}{2}\cdot\dfrac{2^2-1+1}{\left(2-1\right)\left(2+1\right)}\cdot\dfrac{3^2-1+1}{\left(3-1\right)\left(3+1\right)}\cdot...\cdot\dfrac{2016^2-1+1}{\left(2016-1\right)\left(2016+1\right)}\)

\(=\dfrac{1}{2}\cdot\dfrac{2}{1}\cdot\dfrac{3}{2}\cdot...\cdot\dfrac{2016}{2015}\cdot\dfrac{2}{3}\cdot\dfrac{3}{4}\cdot...\cdot\dfrac{2016}{2017}\)

\(=\dfrac{1}{2}\cdot2016\cdot\dfrac{2}{2017}=\dfrac{2016}{2017}\)

P = (1 + \(\dfrac{1}{1.3}\)).(1 + \(\dfrac{1}{2.4}\)).(1 + \(\dfrac{1}{3.5}\))....(1 + \(\dfrac{1}{2020.2022}\))

P = \(\dfrac{1.3+1}{1.3}\). \(\dfrac{2.4+1}{2.4}\).\(\dfrac{3.5+1}{3.5}\)....\(\dfrac{2020.2022+1}{2020.2022}\)

P=\(\dfrac{\left(2-1\right)\left(2+1\right)+1}{1.3}\).\(\dfrac{\left(3-1\right)\left(3+1\right)+1}{2.4}\)...\(\dfrac{\left(2021+1\right).\left(2022-1\right)+1}{2020.2022}\)

P = \(\dfrac{2.2}{1.3}\).\(\dfrac{3.3}{2.4}\).\(\dfrac{4.4}{3.5}\)....\(\dfrac{2021.2021}{2020.2022}\)

P = \(\dfrac{2.2021}{2022}\)

P = \(\dfrac{2021}{1011}\) 

P = (1+\(\dfrac{1}{1.3}\)).(1+\(\dfrac{1}{2.4}\)).(1 + \(\dfrac{1}{3.5}\))...(1+\(\dfrac{1}{2020.2022}\))

P =\(\dfrac{1.3+1}{1.3}\).\(\dfrac{2.4+1}{2.4}\).\(\dfrac{3.5+1}{3.5}\)...\(\dfrac{2020.2022+1}{2020.2022}\)

P = \(\dfrac{(2-1)(2+1)+1}{1.3}\).\(\dfrac{(3-1)(3+1)+1}{2.4}\)...\(\dfrac{(2021-1)(2021+1)}{2020.2022}\)

P = \(\dfrac{2.2}{1.3}\).\(\dfrac{3.3}{2.4}\).\(\dfrac{4.4}{3.5}\)...\(\dfrac{2021.2021}{2020.2022}\)

P = \(\dfrac{2021}{1011}\)

\(\left(1+\dfrac{1}{1.3}\right).\left(1+\dfrac{1}{2.4}\right).\left(1+\dfrac{1}{3.5}\right).........\left[1+\dfrac{1}{x.\left(x+2\right)}\right]=\dfrac{31}{16}\)

\(\Rightarrow\dfrac{2^2}{1.3}.\dfrac{3^2}{2.4}.\dfrac{4^2}{3.5}........\dfrac{\left(x+1\right)^2}{x.\left(x+2\right)}=\dfrac{31}{16}\)

\(\Rightarrow\dfrac{\left[2.3.4.............\left(x+1\right)\right].\left[2.3.4.............\left(x+1\right)\right]}{\left(1.2.3...................x\right).\left(3.4.5..........................\left(x+2\right)\right)}=\dfrac{31}{16}\)

\(\Rightarrow\dfrac{\left(x+1\right).2}{1.\left(x+2\right)}=\dfrac{31}{16}\)

\(\Leftrightarrow16.2\left(x+1\right)=31.\left(x+2\right)\)

\(\Rightarrow32x+32=31x+62\)

\(\Rightarrow x=30\)

Vậy x=30

Chúc bn học tốt

ĐKXĐ: \(x\notin\left\{0;-2\right\}\)

Ta có: \(\left(1+\dfrac{1}{1\cdot3}\right)\left(1+\dfrac{1}{2\cdot4}\right)\left(1+\dfrac{1}{3\cdot5}\right)\cdot...\cdot\left(1+\dfrac{1}{x\left(x+2\right)}\right)=\dfrac{31}{16}\)

\(\Leftrightarrow\dfrac{1\cdot3+1}{1\cdot3}+\dfrac{1+2\cdot4}{2\cdot4}+\dfrac{1+3\cdot5}{3\cdot5}\cdot...\cdot\dfrac{1+x\left(x+2\right)}{x\left(x+2\right)}=\dfrac{31}{16}\)

\(\Leftrightarrow\dfrac{2\cdot2}{1\cdot3}+\dfrac{3\cdot3}{2\cdot4}+\dfrac{4\cdot4}{3\cdot5}+...+\dfrac{\left(x+1\right)\left(x+1\right)}{x\left(x+2\right)}=\dfrac{31}{16}\)

\(\Leftrightarrow\dfrac{1\cdot2\cdot3\cdot...\cdot\left(x+1\right)}{1\cdot2\cdot3\cdot...\cdot x}\cdot\dfrac{2\cdot3\cdot4\cdot...\cdot\left(x+1\right)}{3\cdot4\cdot5\cdot...\cdot\left(x+2\right)}=\dfrac{31}{16}\)

\(\Leftrightarrow\left(x+1\right)\cdot\dfrac{2}{x+2}=\dfrac{31}{16}\)

\(\Leftrightarrow\dfrac{2x+2}{x+2}=\dfrac{31}{16}\)

\(\Leftrightarrow\dfrac{32x+32}{16\left(x+2\right)}=\dfrac{31\left(x+2\right)}{16\left(x+2\right)}\)

Suy ra: \(32x+32=31x+62\)

\(\Leftrightarrow x=30\)(thỏa ĐK)

Vậy: S={30}