Bài 1:
a) \(\dfrac{53}{101}.\dfrac{-13}{97}+\dfrac{53}{101}.\dfrac{-84}{97}\)
\(=\dfrac{53}{101}.\left(\dfrac{-13}{97}+\dfrac{-84}{97}\right)\)
\(=\dfrac{53}{101}.\left(-1\right)\)
\(=\dfrac{-53}{101}\)
Bài 1:
a) Ta có: \(\dfrac{53}{101}\cdot\dfrac{-13}{97}+\dfrac{53}{101}\cdot\dfrac{-84}{97}\)
\(=\dfrac{53}{101}\left(-\dfrac{13}{97}-\dfrac{84}{97}\right)\)
\(=\dfrac{53}{101}\cdot\left(-1\right)=\dfrac{-53}{101}\)
b) Ta có: \(\left(\dfrac{1}{57}-\dfrac{1}{5757}+\dfrac{1}{23}\right)\left(\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{6}\right)\)
\(=\left(\dfrac{1}{57}-\dfrac{1}{5757}+\dfrac{1}{23}\right)\left(\dfrac{3}{6}-\dfrac{2}{6}-\dfrac{1}{6}\right)\)
=0
Bài 1:
c) Ta có: \(\left(1+\dfrac{1}{2}\right)\left(1+\dfrac{1}{3}\right)\left(1+\dfrac{1}{4}\right)\cdot...\cdot\left(1+\dfrac{1}{2020}\right)\left(1+\dfrac{1}{2021}\right)\)
\(=\dfrac{3}{2}\cdot\dfrac{4}{3}\cdot\dfrac{5}{4}\cdot...\cdot\dfrac{2021}{2020}\cdot\dfrac{2022}{2021}\)
\(=\dfrac{2022}{2}=1011\)
d) Ta có: \(\left(1-\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)\cdot...\cdot\left(1-\dfrac{1}{199}\right)\left(1-\dfrac{1}{200}\right)\)
\(=\dfrac{-1}{2}\cdot\dfrac{-2}{3}\cdot...\cdot\dfrac{-198}{199}\cdot\dfrac{-199}{200}\)
\(=-\left(\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot...\cdot\dfrac{198}{199}\cdot\dfrac{199}{200}\right)\)
\(=-\dfrac{1}{200}\)
Bài 3:
d) Ta có: \(\dfrac{-1}{999\cdot1000}-\dfrac{1}{998\cdot999}-\dfrac{1}{997\cdot998}-...-\dfrac{1}{1\cdot2}\)
\(=-\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{997\cdot998}+\dfrac{1}{998\cdot999}+\dfrac{1}{999\cdot1000}\right)\)
\(=-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{998}-\dfrac{1}{999}+\dfrac{1}{999}-\dfrac{1}{1000}\right)\)
\(=-\left(1-\dfrac{1}{1000}\right)\)
\(=-\dfrac{999}{1000}\)
Bài 1:
a) \(\dfrac{53}{101}.\dfrac{-13}{97}+\dfrac{53}{101}.\dfrac{-84}{97}\)
\(=\dfrac{53}{101}.\left(\dfrac{-13}{97}+\dfrac{-84}{97}\right)\)
\(=\dfrac{53}{101}.-1\)
\(=\dfrac{-53}{101}\)
b) \(\left(\dfrac{1}{57}-\dfrac{1}{5757}+\dfrac{1}{23}\right).\left(\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{6}\right)\)
\(=\left(\dfrac{1}{57}-\dfrac{1}{5757}+\dfrac{1}{23}\right).0\)
\(=0\)
c) \(\left(1+\dfrac{1}{2}\right).\left(1+\dfrac{1}{3}\right).\left(1+\dfrac{1}{4}\right).....\left(1+\dfrac{1}{2020}\right).\left(1+\dfrac{1}{2021}\right)\)
\(=\dfrac{3}{2}.\dfrac{4}{3}.\dfrac{5}{4}.....\dfrac{2021}{2020}.\dfrac{2022}{2021}\)
\(=\dfrac{\left(3.4.5.....2021.2022\right)}{\left(2.3.4.....2020.2021\right)}\)
\(=\dfrac{2022}{2}=1011\)
d) \(\left(1-\dfrac{1}{2}\right).\left(1-\dfrac{1}{3}\right).....\left(1-\dfrac{1}{199}\right).\left(1-\dfrac{1}{200}\right)\)
\(=\dfrac{1}{2}.\dfrac{2}{3}.\dfrac{3}{4}.....\dfrac{198}{199}.\dfrac{199}{200}\)
\(=\dfrac{\left(1.2.3.....198.199\right)}{\left(2.3.4.....199.200\right)}\)
\(=\dfrac{1}{200}\)
Bài 2:
a) \(\dfrac{1^2}{1.2}.\dfrac{2^2}{2.3}.\dfrac{3^2}{3.4}.....\dfrac{99^2}{99.100}.\dfrac{100^2}{100.101}\)
\(=\dfrac{1.1.2.2.3.3.....99.99.100.100}{1.2.2.3.3.4.....99.100.100.101}\)
\(=\dfrac{1}{1.101}=\dfrac{1}{101}\)
b) \(\dfrac{2^2}{1.3}.\dfrac{3^2}{2.4}.\dfrac{4^2}{3.5}.....\dfrac{59^2}{58.60}\)
\(=\dfrac{2.2.3.3.4.4.....59.59}{1.3.2.4.3.5.....58.60}\)
\(=\dfrac{\left(2.3.4.....59\right)}{\left(1.2.3.....58\right)}.\dfrac{\left(2.3.4.....59\right)}{\left(3.4.5.....60\right)}\)
\(=\dfrac{59}{1}.\dfrac{2}{60}\)
\(=59.\dfrac{1}{30}\)
\(=\dfrac{59}{30}\)
Bài 3:
a) \(\left(\dfrac{1}{24}-\dfrac{5}{16}\right):\dfrac{-3}{8}+1=\dfrac{-13}{48}:\dfrac{-3}{8}+1=\dfrac{13}{18}+1=\dfrac{31}{18}\)
b) \(\dfrac{-5}{7}:\left(25-24\dfrac{4}{7}\right)+15\dfrac{1}{3}.\dfrac{-1}{23}\)
\(=\dfrac{-5}{7}:\left(25-\dfrac{172}{7}\right)+\dfrac{46}{3}.\dfrac{-1}{23}\)
\(=\dfrac{-5}{7}:\dfrac{3}{7}+\dfrac{-2}{3}\)
\(=\dfrac{-5}{3}+\dfrac{-2}{3}\)
\(=\dfrac{-7}{3}\)
d) \(-\dfrac{1}{999.1000}-\dfrac{1}{998.999}-\dfrac{1}{997.998}-...-\dfrac{1}{1.2}\)
\(=-1.\left(\dfrac{1}{1.2}+...+\dfrac{1}{997.998}+\dfrac{1}{998.999}+\dfrac{1}{999.1000}\right)\)
\(=-1.\left(\dfrac{1}{1}-\dfrac{1}{2}+...+\dfrac{1}{997}-\dfrac{1}{998}+\dfrac{1}{998}-\dfrac{1}{999}+\dfrac{1}{999}-\dfrac{1}{1000}\right)\)
\(=-1.\left(\dfrac{1}{1}-\dfrac{1}{1000}\right)\)
\(=-1.\dfrac{999}{1000}=\dfrac{-999}{1000}\)
Xin lỗi bạn vì mk ko bt làm câu c! 
