Ta có : \(P=1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{2019}-\dfrac{1}{2020}=1-\dfrac{1}{2020}=\dfrac{2019}{2020}\)

mà \(2019< 2020\)nên P < 1 ( đpcm ) 

\(P=\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{2019.2021}\) 

\(P=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{2019}-\dfrac{1}{2021}\) 

\(P=1-\dfrac{1}{2021}\) 

\(P=\dfrac{2020}{2021}\)

Vì \(\dfrac{2020}{2021}< 1\) ⇒ \(P< 1\) ( điều phải chứng minh ) 

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

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

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

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

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

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

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

Lời giải:
Gọi tích trên là $A$

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

Thay $n=1,2,3....,2019$ ta có:

$A=\frac{2^2}{1.3}.\frac{3^2}{2.4}.\frac{4^2}{3.5}....\frac{2020^2}{2019.2021}$

$=\frac{2^2.3^2...2020^2}{(1.3)(2.4)(3.5)...(2019.2021)}$

$=\frac{(2.3....2020)(2.3...2020)}{(1.2.3...2019)(3.4...2021)}$

$=2020.\frac{2}{2021}=\frac{4040}{2021}$

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

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

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

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

\(=\frac{2^2\cdot3^2\cdot\ldots\cdot2020^2}{\left(1\cdot3\right)\cdot\left(2\cdot4\right)\cdot\ldots\cdot\left(2019\cdot2021\right)}=\frac{2\cdot3\cdot\ldots\cdot2020}{1\cdot2\cdot\ldots\cdot2019}\cdot\frac{2\cdot3\cdot\ldots\cdot2020}{3\cdot4\cdot\ldots\cdot2021}\)

\(=\frac{2020}{1}\cdot\frac{2}{2021}=\frac{4040}{2021}\)

Sửa đề: A=(1+1/1*3)(1+1/2*4)*...*(1+1/2019*2021)

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

\(=\dfrac{2}{1}\cdot\dfrac{3}{2}\cdot...\cdot\dfrac{2020}{2019}\cdot\dfrac{2}{3}\cdot\dfrac{3}{4}\cdot...\cdot\dfrac{2020}{2021}=2020\cdot\dfrac{2}{2021}=\dfrac{4040}{2021}\)

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

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

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

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

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

\(=\frac{2\cdot2}{1\cdot3}\cdot\frac{3\cdot3}{2\cdot4}\cdot\ldots\cdot\frac{2020\cdot2020}{2019\cdot2021}\)

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

   1/1x3 + 1/3x5 + 1/5x7 + ...............................+ 1/97x99

=1-1/3 + 1/3 - 1/5 + 1/5 - 1/7 +.............................+ 1/97-1/99

=1-1/99

=98/99

Đặt A=1/1*3+1/3*5+..+1/99*101

A=2/2*(1/1*3+1/3*5+...+1/99*101)

A=1/2*(2/1*3+2/3*5+..+2/99*101)

A=1/2*(1/1-1/3+1/3-1/5+...+1/99-1/100)

A=1/2*(1/1-1/100)

A=1/2*99/100

A=99/200

50/101 nha

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