Bài 11:
Ta có: \(A=\left(1-\frac12\right)\left(1-\frac13\right)\cdot\ldots\cdot\left(1-\frac{1}{2021}\right)\)
\(=\frac12\cdot\frac23\cdot\ldots\cdot\frac{2020}{2021}\)
\(=\frac{1}{2021}\)
Câu 12: \(A=\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)\cdot\ldots\cdot\left(1-\frac{1}{100^2}\right)\)
\(=\left(1-\frac12\right)\left(1-\frac13\right)\cdot\ldots\cdot\left(1-\frac{1}{100}\right)\left(1+\frac12\right)\left(1+\frac13\right)\cdot\ldots\cdot\left(1+\frac{1}{100}\right)\)
\(=\frac12\cdot\frac23\cdot\ldots\cdot\frac{99}{100}\cdot\frac32\cdot\frac43\cdot\ldots\cdot\frac{101}{100}=\frac{1}{100}\cdot\frac{101}{2}=\frac{101}{200}\)
gấp gấp cứu e ạ
