Bạn xem bài tương tự tại đây. Đề là:
Tính $(1+\frac{1}{1.3})(1+\frac{1}{2.4})....(1+\frac{1}{2021.2023})$

Sửa đề: \(a=\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+\cdots+\frac{1}{2019\cdot2021}+\frac{1}{2021\cdot2023}\)

Ta có: \(a=\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+\cdots+\frac{1}{2019\cdot2021}+\frac{1}{2021\cdot2023}\)

\(=\frac12\left(\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\cdots+\frac{2}{2019\cdot2021}+\frac{2}{2021\cdot2023}\right)\)

\(=\frac12\left(1-\frac13+\frac13-\frac15+\cdots+\frac{1}{2019}-\frac{1}{2021}+\frac{1}{2021}-\frac{1}{2023}\right)\)

\(=\frac12\left(1-\frac{1}{2023}\right)=\frac12\cdot\frac{2022}{2023}=\frac{1011}{2023}\)

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

câu này khó thế

cong nhan

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

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

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

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

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

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

mk cx co dap an vay

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

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

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

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

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

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

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

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