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 ) 

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

= \(2.\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+...+\dfrac{2}{2019.2021}\right)\)

= \(1.\left(\dfrac{1}{1.3}+\dfrac{1}{3.5}+...+\dfrac{1}{2019.2021}\right)\)

= \(1.\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{2019}-\dfrac{1}{2021}\right)\)

= \(1.\left(1-\dfrac{1}{2021}\right)\)

= \(1.\dfrac{2020}{2021}\)

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

làm cho mn câu b đi

Đặt \(A=\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+...+\frac{1}{2019\cdot2021}\)

\(2A=\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+....+\frac{2}{2019\cdot2021}\)

\(2A=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+....+\frac{1}{2019}-\frac{1}{2021}\)

\(2A=1-\frac{1}{2021}=\frac{2020}{2021}\)

\(A=\frac{2020}{2021}:2=\frac{2020\cdot2}{2021}=\frac{4040}{2021}\)

bn 

Tran Le Khanh Linh lm sai r nếu chia 2 thì 2021.2 chứ ko phải 2020.2

A=2 + 2^2+ 2^3 + 2^4 + 2^5 + 2^6 + 2^6 + 2^7 + 2^8 + 2^9

=>A=(2+2^2+2^3)+....+(2^7+2^8+2^9)

A=(2.1+2.2+2.2^2)+.......+(2^7.1+2^7.2+2^7.2^2)

A=2(1+2+2^2)+....+2^7(1+2+2^2)

A=2.7+....+2^7.7

A=(2+...+2^7).7 chia hết cho 7

vậy A chia hết cho 7

tick nha

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

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

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

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

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

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

B=1-1/4+1/4-1/7+1/7-1/10+...+1/27-1/30.                                                                                                                                              =>B=1-1/30=29/30.

=1/1-1/4+1/4-1/7+1/7-1/10+...+1/27-1/30

=1/1-(1/4-1/4)-(1/7-1/7)-(1/10-1/10)-...-(1/27-1/27)-1/30

 =1-0-0-0-...-0-1/30

=1-1/30

=1/29

  Bạn tự kết luận nhé..!

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

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

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

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

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

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

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

=2.(2\1.3+2\3.5+...+2\9.11)

=2.(1-1\11)

làm tắt bạn tự hiểu nhé