Ta có :

\(A=\left(1+\dfrac{1}{1.3}\right)\left(1+\dfrac{1}{2.4}\right)\left(1+\dfrac{1}{3.5}\right).................\left(1+\dfrac{1}{2011.2013}\right)\)

\(A=\left(\dfrac{3}{3}+\dfrac{1}{3}\right)\left(\dfrac{8}{8}+\dfrac{1}{8}\right)................\left(\dfrac{9999}{9999}+\dfrac{1}{9999}\right)\)

\(A=\dfrac{4}{3}.\dfrac{9}{8}...............\dfrac{10000}{9999}\)

\(A=\dfrac{4.9.................10000}{3.8.............9999}\)

\(A=\dfrac{2.2.3.3................100.100}{1.3.2.4...............99.101}\)

\(A=\dfrac{2.100}{101}=\dfrac{200}{101}\)

~ Chúc bn học tốt ~

A= (1+1/1 x 3)x(1+1/2x4)x(1+1/3x5)x............x(1+1/2011x2013)

\(=\left(\frac{3}{3}+\frac{1}{3}\right)\left(\frac{8}{8}+\frac{1}{8}\right)....\left(\frac{4048143}{4048143}+\frac{1}{4048143}\right)\)

\(=\frac{4}{3}\cdot\frac{9}{8}\cdot...\cdot\frac{4048144}{4048143}\)

\(=\frac{4\cdot9\cdot....\cdot4048144}{3\cdot8\cdot....\cdot4048143}\)

\(=\frac{2\cdot2\cdot3\cdot3\cdot....\cdot2012\cdot2012}{1\cdot3\cdot2\cdot4\cdot....\cdot2011\cdot2013}\)

\(=\frac{2\cdot2012}{2013}=\frac{4024}{2013}\)

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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