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

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

Giải:

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

\(\Leftrightarrow A=\dfrac{1}{2}.\dfrac{2}{3}...\dfrac{2015}{2016}.\dfrac{2016}{2017}\)

\(\Leftrightarrow A=\dfrac{1.2...201.2016}{2.3...2016.2017}\)

\(\Leftrightarrow A=\dfrac{1.2.3...2015.2016}{2017.2.3...2015.2016.}\)

Rút gọ cả tử và mẫu với 2.3...2015.2016, ta được:

\(A=\dfrac{1}{2017}\)

Vậy \(A=\dfrac{1}{2017}\).

Chúc bạn học tốt!

Giải:

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

\(A=\dfrac{1}{2017}\)

Vậy, A= \(\dfrac{1}{2017}\)

\(=\dfrac{1}{99\cdot97}-\left(\dfrac{1}{1\cdot3}+\dfrac{1}{3\cdot5}+...+\dfrac{1}{95\cdot97}\right)\)

\(=\dfrac{1}{99\cdot97}-\dfrac{1}{2}\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{95\cdot97}\right)\)

\(=\dfrac{1}{99\cdot97}-\dfrac{1}{2}\cdot\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{95}-\dfrac{1}{97}\right)\)

\(=\dfrac{1}{99\cdot97}-\dfrac{48}{97}=\dfrac{1-48\cdot99}{97\cdot99}=\dfrac{-4751}{9603}\)

\(Q=\dfrac{-2015}{2016}\cdot\left(-50\right)\cdot\dfrac{-153}{154}\cdot1\dfrac{1}{2015}\cdot20\%\)

\(=\dfrac{-2015}{2016}\cdot\left(-50\right)\cdot\dfrac{-153}{154}\cdot\dfrac{2016}{2015}\cdot\dfrac{1}{5}\\ =\left(-\dfrac{2015}{2016}\cdot\dfrac{2016}{2015}\right)\cdot\left(-50\cdot\dfrac{1}{5}\right)\cdot-\dfrac{153}{154}\\ =\left(-1\right)\cdot\left(-10\right)\cdot\left(-\dfrac{153}{154}\right)\\ =10\cdot\left(-\dfrac{153}{154}\right)\\ =-\dfrac{1530}{154}\\ =-\dfrac{765}{77}\)

\(A=\dfrac{1}{99.100}-\dfrac{1}{98.99}-....-\dfrac{1}{3.2}-\dfrac{1}{2.1}\\ =-\left(-\dfrac{1}{99.100}+\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{98.99}\right)\\ =-\left(-\dfrac{1}{99.100}+\dfrac{98}{99}\right)\\ =\dfrac{1}{99.100}-\dfrac{98}{99}\\ =\dfrac{1}{99}\left(\dfrac{1}{100}-98\right)=\dfrac{-9799}{9900}\)

\(C=\dfrac{1}{100}-\dfrac{1}{100\cdot99}-\dfrac{1}{99\cdot98}-\dfrac{1}{98\cdot97}-...-\dfrac{1}{3\cdot2}-\dfrac{1}{2\cdot1}\)

\(C=\dfrac{1}{100}-\left(\dfrac{1}{2\cdot1}+\dfrac{1}{3\cdot2}+...+\dfrac{1}{98\cdot99}+\dfrac{1}{99\cdot100}\right)\)

\(C=\dfrac{1}{100}-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{98}-\dfrac{1}{99}+\dfrac{1}{99}-\dfrac{1}{100}\right)\)

\(C=\dfrac{1}{100}-\left(1-\dfrac{1}{100}\right)\)

\(C=\dfrac{1}{100}-\dfrac{99}{100}=\dfrac{-98}{100}=-\dfrac{49}{50}\)

\(C=\frac{1}{100}-\frac{1}{100.99}-\frac{1}{99.98}-...-\frac{1}{3.2}-\frac{1}{2.1}\)

\(C=\frac{1}{100}-\left(\frac{1}{1.2}+\frac{1}{2.3}+..+\frac{1}{98.99}+\frac{1}{99.100}\right)\)

\(C=\frac{1}{100}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\right)\)

\(C=\frac{1}{100}-\left(1-\frac{1}{100}\right)\)

\(C=\frac{1}{100}-\frac{99}{100}\)

\(C=-\frac{98}{100}=-\frac{49}{50}\)

Sửa đề: \(\dfrac{1}{99.97}-\dfrac{1}{97.95}-\dfrac{1}{95.93}-...-\dfrac{1}{3.1}\)

\(=\dfrac{1}{97.99}-\left(\dfrac{1}{1.3}+...+\dfrac{1}{93.95}+\dfrac{1}{95.97}\right)\)

\(=\dfrac{1}{97.99}-\dfrac{1}{2}\left(\dfrac{2}{1.3}+...+\dfrac{2}{93.95}+\dfrac{2}{95.97}\right)\)

\(=\dfrac{1}{97.99}-\dfrac{1}{2}\left(1-\dfrac{1}{3}+...+\dfrac{1}{93}-\dfrac{1}{95}+\dfrac{1}{95}-\dfrac{1}{97}\right)\)

\(=\dfrac{1}{97.99}-\dfrac{1}{2}\left(1-\dfrac{1}{97}\right)\)

\(=\dfrac{1}{2}\left(\dfrac{2}{97.99}\right)-\dfrac{1}{2}.\dfrac{96}{97}\)

\(=\dfrac{1}{2}\left(\dfrac{1}{97}-\dfrac{1}{99}\right)-\dfrac{48}{97}\)

.........................

Đã nói là 5ps mà Tú nếu có ... thì tui làm đc rồi

Bạn Tú giải đúng r tuy vio ko có"..."nhưng tui khẳng định có mọi người đừng bị lừa như tôi nha(tôi nhập đáp án \(-\dfrac{4751}{9603}\) thì đúng tui khẳng định một lần nữa là có"..." huhu)

please help me

\(\dfrac{1+\dfrac{1}{3}+\dfrac{1}{5}+\dfrac{1}{7} +.....................+\dfrac{1}{97}+\dfrac{1}{99}}{\dfrac{1}{1.99}+\dfrac{1}{3.97}+\dfrac{1}{5.95}+....+\dfrac{1}{97.3}+\dfrac{1}{99.1}}\)

\(=\dfrac{\left(1+\dfrac{1}{99}\right)+\left(\dfrac{1}{3}+\dfrac{1}{97}\right)+..........+\left(\dfrac{1}{49}+\dfrac{1}{51}\right)}{2\left(\dfrac{1}{1.99}+\dfrac{1}{3.97}+.......+\dfrac{1}{49.51}\right)}\)

\(=\dfrac{\dfrac{100}{1.99}+\dfrac{100}{3.97}+...........+\dfrac{100}{49.51}}{2\left(\dfrac{1}{1.99}+\dfrac{1}{3.97}+...........+\dfrac{1}{49.51}\right)}\)

\(=\dfrac{100\left(\dfrac{1}{1.99}+\dfrac{1}{3.97}+.............+\dfrac{1}{49.51}\right)}{2\left(\dfrac{1}{1.99}+\dfrac{1}{3.97}+..........+\dfrac{1}{49.51}\right)}\)

\(=\dfrac{100}{2}\)

\(=50\)

\(\dfrac{1+\dfrac{1}{3}+\dfrac{1}{5}+\dfrac{1}{7}+.....+\dfrac{1}{97}+\dfrac{1}{99}}{\dfrac{1}{1.99}+\dfrac{1}{3.97}+\dfrac{1}{5.95}+...+\dfrac{1}{97.3}+\dfrac{1}{99.1}}=\dfrac{\left(1+\dfrac{1}{99}\right)+\left(\dfrac{1}{3}+\dfrac{1}{97}\right)+....+\left(\dfrac{1}{49}+\dfrac{1}{51}\right)}{2\left(\dfrac{1}{1.99}+\dfrac{1}{3.97}+\dfrac{1}{5.95}+.....+\dfrac{1}{49.51}\right)}=\dfrac{\dfrac{100}{99}+\dfrac{100}{3.97}+....+\dfrac{100}{49.51}}{2\left(\dfrac{1}{1.99}+\dfrac{1}{3.97}+....+\dfrac{1}{49.51}\right)}=\dfrac{100}{2}=50\)

1<1/3+1/8+1/15+1/24+....+1/360>

KO BIẾT ĐÚNG HAY KO NHÉ BẠN 

\(1\frac{1}{3}\)là hỗn số mà

\(1\frac{1}{3}.1\frac{1}{8}.1\frac{1}{15}.1\frac{1}{24}...1\frac{1}{360}\)

\(=\frac{4}{3}.\frac{9}{8}.\frac{16}{15}.\frac{25}{24}...\frac{361}{360}\)

\(=\frac{2^2}{3}.\frac{3^2}{8}.\frac{4^2}{15}.\frac{5^2}{24}...\frac{19^2}{360}\)

\(=\frac{2.2}{1.3}.\frac{3.3}{2.4}.\frac{4.4}{3.5}.\frac{5.5}{4.6}...\frac{19.19}{18.20}\)

\(=\left(\frac{2.3.4.5...19}{1.2.3.4...18}\right).\left(\frac{2.3.4.5...19}{3.4.5.6...20}\right)\)

\(=19.\frac{1}{10}\)

\(=\frac{19}{10}\)