Question

[Takitori]

Tính \(B=\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).\left(1-\frac{1}{4}\right)...\left(1-\frac{1}{20}\right)\)

Answer 1

[Lê Tài Bảo Châu]

\(B=\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).....\left(1-\frac{1}{20}\right)\)

\(B=\frac{1}{2}.\frac{2}{3}.....\frac{19}{20}\)

\(B=\frac{1}{20}\)

Answer 2

[nguyen duc thang]

B = \(\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).\left(1-\frac{1}{4}\right)...\left(1-\frac{1}{20}\right)\)

B = \(\frac{1}{2}.\frac{2}{3}.\frac{3}{4}...\frac{19}{20}\)'

B = \(\frac{1}{20}\)

Vậy B = \(\frac{1}{20}\)