Ta có: \(\left(1+\frac{1}{1\cdot3}\right)\left(1+\frac{1}{2\cdot4}\right)\cdot\ldots\cdot\left(1+\frac{1}{2019\cdot2021}\right)\)

\(=\left(1+\frac{1}{\left(2-1\right)\left(2+1\right)}\right)\left(1+\frac{1}{\left(3-1\right)\left(3+1\right)}\right)\cdot\ldots\cdot\left(1+\frac{1}{\left(2020-1\right)\left(2020+1\right)}\right)\)

\(=\left(1+\frac{1}{2^2-1}\right)\left(1+\frac{1}{3^2-1}\right)\cdot\ldots\cdot\left(1+\frac{1}{2020^2-1}\right)\)

\(=\frac{2^2-1+1}{2^2-1}\cdot\frac{3^2-1+1}{3^2-1}\cdot\ldots\cdot\frac{2020^2-1+1}{2020^2-1}\)

\(=\frac{2^2}{1\cdot3}\cdot\frac{3^2}{2\cdot4}\cdot\ldots\cdot\frac{2020^2}{2019\cdot2021}\)

\(=\frac{2\cdot3\cdot\ldots\cdot2020}{1\cdot2\cdot\ldots\cdot2019}\cdot\frac{2\cdot3\cdot\ldots\cdot2020}{3\cdot4\cdot\ldots\cdot2021}=\frac{2020}{1}\cdot\frac{2}{2021}=\frac{4040}{2021}\)

Lời giải:
Xét thừa số tổng quát $1+\frac{1}{n(n+2)}=\frac{n(n+2)+1}{n(n+2)}=\frac{(n+1)^2}{n(n+2)}$

Khi đó:

$1+\frac{1}{1.3}=\frac{2^2}{1.3}$

$1+\frac{1}{2.4}=\frac{3^2}{2.4}$

.........

$1+\frac{1}{99.101}=\frac{100^2}{99.101}$

Khi đó:

$A=\frac{2^2.3^2.4^2......100^2}{(1.3).(2.4).(3.5)....(99.101)}$

$=\frac{(2.3.4...100)(2.3.4...100)}{(1.2.3...99)(3.4.5...101)}$

$=\frac{2.3.4...100}{1.2.3..99}.\frac{2.3.4...100}{3.4.5..101}$
$=100.\frac{2}{101}=\frac{200}{101}$

giúp em với

\(\left(1+\frac{1}{1\times3}\right)\times\left(1+\frac{1}{2\times4}\right)\times\left(1+\frac{1}{3\times5}\right)\times...\times\left(1+\frac{1}{99.101}\right)\)

\(=\left(\frac{3}{3}+\frac{1}{3}\right)\times\left(\frac{8}{8}+\frac{1}{8}\right)\times\left(\frac{15}{15}+\frac{1}{15}\right)\times...\times\left(\frac{9999}{9999}+\frac{1}{9999}\right)\)

\(=\frac{4}{3}\times\frac{9}{8}\times\frac{16}{15}\times...\times\frac{10000}{9999}\)

\(=\frac{4\times9\times16\times...\times10000}{3\times8\times15\times...\times9999}\)

\(=\frac{2\times2\times3\times3\times4\times4\times...\times100\times100}{1\times3\times2\times4\times3\times5\times...\times99\times101}\)

\(=\frac{2\times100}{101}=\frac{200}{101}\)

mk cx co dap an vay

Ta có: \(\left(1+\frac{1}{1\cdot3}\right)\left(1+\frac{1}{2\cdot4}\right)\cdot\ldots\cdot\left(1+\frac{1}{99\cdot101}\right)\cdot x=\frac{100}{101}\)

=>\(\left(1+\frac{1}{2^2-1}\right)\left(1+\frac{1}{3^2-1}\right)\cdot\ldots\cdot\left(1+\frac{1}{100^2-1}\right)\cdot x=\frac{100}{101}\)

=>\(\frac{2^2}{2^2-1}\cdot\frac{3^2}{3^2-1}\cdot\ldots\cdot\frac{100^2}{100^2-1}\cdot x=\frac{100}{101}\)

=>\(\frac{2^2}{1\cdot3}\cdot\frac{3^2}{2\cdot4}\cdot\ldots\cdot\frac{100^2}{99\cdot101}\cdot x=\frac{100}{101}\)

=>\(\frac{2\cdot3\cdot\ldots\cdot100}{1\cdot2\cdot3\cdot\ldots\cdot99}\cdot\frac{2\cdot3\cdot\ldots\cdot100}{3\cdot4\cdot\ldots\cdot101}\cdot x=\frac{100}{101}\)

=>\(100\cdot\frac{2}{101}\cdot x=\frac{100}{101}\)

=>\(\frac{200}{101}\cdot x=\frac{100}{101}\)

=>\(x=\frac{100}{101}:\frac{200}{101}=\frac{100}{200}=\frac12\)

câu này khó thế

cong nhan

Ta có: \(A=\frac12\cdot\left(1+\frac{1}{1\cdot3}\right)\left(1+\frac{1}{2\cdot4}\right)\cdot\ldots\cdot\left(1+\frac{1}{2021\cdot2023}\right)\)

\(=\frac12\left(1+\frac{1}{2^2-1}\right)\left(1+\frac{1}{3^2-1}\right)\cdot\ldots\cdot\left(1+\frac{1}{2022^2-1}\right)\)

\(=\frac12\cdot\frac{2^2-1+1}{2^2-1}\cdot\frac{3^2-1+1}{3^2-1}\cdot\ldots\cdot\frac{2022^2-1+1}{2022^2-1}\)

\(=\frac12\cdot\frac{2^2}{2^2-1}\cdot\frac{3^2}{3^2-1}\cdot\ldots\cdot\frac{2022^2}{2022^2-1}\)

\(=\frac12\cdot\frac{2^2}{1\cdot3}\cdot\frac{3^2}{2\cdot4}\cdot\ldots\cdot\frac{2022^2}{2021\cdot2023}=\frac12\cdot\frac{2\cdot3\cdot\ldots\cdot2022}{1\cdot2\cdot\ldots\cdot2021}\cdot\frac{2\cdot3\cdot\ldots\cdot2022}{3\cdot4\cdot\ldots\cdot2023}\)

\(=\frac12\cdot2022\cdot\frac{2}{2023}=\frac{2022}{2023}\)