Bài 2:
a: \(\dfrac{1^2}{1\cdot2}\cdot\dfrac{2^2}{2\cdot3}\cdot...\cdot\dfrac{99^2}{99\cdot100}\cdot\dfrac{100^2}{100\cdot101}\)
\(=\dfrac{1\cdot2\cdot3\cdot...\cdot99\cdot100}{1\cdot2\cdot3...\cdot99\cdot100}\cdot\dfrac{1\cdot2\cdot3\cdot...\cdot99\cdot100}{2\cdot3\cdot...\cdot100\cdot101}\)
\(=1\cdot\dfrac{1}{101}=\dfrac{1}{101}\)
b: \(\dfrac{2^2}{1\cdot3}\cdot\dfrac{3^2}{2\cdot4}\cdot\dfrac{4^2}{3\cdot5}\cdot...\cdot\dfrac{59^2}{58\cdot60}\)
\(=\dfrac{2\cdot3\cdot4\cdot...\cdot59}{1\cdot2\cdot3\cdot...\cdot58}\cdot\dfrac{2\cdot3\cdot4\cdot...\cdot59}{3\cdot4\cdot5\cdot...\cdot59\cdot60}\)
\(=\dfrac{59}{1}\cdot\dfrac{2}{60}=\dfrac{59}{30}\)
Bài 1:
a: \(\dfrac{53}{101}\cdot\dfrac{-13}{97}+\dfrac{53}{101}\cdot\dfrac{-84}{97}\)
\(=\dfrac{53}{101}\left(-\dfrac{13}{97}-\dfrac{84}{97}\right)\)
\(=\dfrac{53}{101}\cdot\left(-1\right)=-\dfrac{53}{101}\)
b: \(\left(\dfrac{1}{57}-\dfrac{1}{5757}+\dfrac{1}{23}\right)\cdot\left(\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{6}\right)\)
\(=\left(\dfrac{1}{57}-\dfrac{1}{5757}+\dfrac{1}{23}\right)\cdot\dfrac{3-2-1}{6}\)
=0
c: \(\left(1+\dfrac{1}{2}\right)\left(1+\dfrac{1}{3}\right)\cdot...\cdot\left(1+\dfrac{1}{2020}\right)\left(1+\dfrac{1}{2021}\right)\)
\(=\dfrac{3}{2}\cdot\dfrac{4}{3}\cdot...\cdot\dfrac{2021}{2020}\cdot\dfrac{2022}{2021}\)
\(=\dfrac{2022}{2}=1011\)
d: \(\left(1-\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)\cdot...\cdot\left(1-\dfrac{1}{199}\right)\left(1-\dfrac{1}{200}\right)\)
\(=\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot...\cdot\dfrac{198}{199}\cdot\dfrac{199}{200}\)
\(=\dfrac{1}{200}\)
