Ta có: \(A=\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{2019\cdot2020}+\dfrac{1}{2020\cdot2021}\)

\(=\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2019}-\dfrac{1}{2020}+\dfrac{1}{2020}-\dfrac{1}{2021}\)

\(=\dfrac{1}{1}-\dfrac{1}{2021}=\dfrac{2021}{2021}-\dfrac{1}{2021}\)

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

mà \(\dfrac{2020}{2021}< \dfrac{2021}{2021}=1\)

nên A<1

\(\left|x+\frac{1}{1\cdot2}\right|+\left|x+\frac{1}{2\cdot3}\right|+\cdots+\left|x+\frac{1}{2020\cdot2021}\right|=2021x\) (1)

=>2021x>=0

=>x>=0

(1) sẽ trở thành:

\(2021x=x+\frac{1}{1\cdot2}+x+\frac{1}{2\cdot3}+\cdots+x+\frac{1}{2020\cdot2021}\)

=>\(2021x=2020x+1-\frac12+\frac12-\frac13+\cdots+\frac{1}{2020}-\frac{1}{2021}\)

=>\(x=1-\frac{1}{2021}=\frac{2020}{2021}\) (nhận)

Ta có: \(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{2020\cdot2021}+\dfrac{1}{2021\cdot2022}\)

\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2020}-\dfrac{1}{2021}+\dfrac{1}{2021}-\dfrac{1}{2022}\)

\(=1-\dfrac{1}{2022}=\dfrac{2021}{2022}\)

1/1x2+1/2x3+1/3x4+...+1/2020x2021+1/2021x2022

=1/1-1/2+1/2-1/3+1/3-1/4+...+1/2020-1/2021+1/2021-1/2022.

=1/1-1/2022

=2021/2022

A = 9/1.2 + 9/2.3 + 9/3.4 + .. + 9/98.99 + 9/99.100
             = 1 - 9/100
             = 100/100 - 9/100
              = 91/100

\(A=9\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{2021}-\dfrac{1}{2022}\right)\)

\(A=9\left(1-\dfrac{1}{2022}\right)=9.\dfrac{2021}{2022}=\dfrac{18189}{2022}=\dfrac{6063}{674}\)

Ta có : S = 1.2 + 2.3 + 3.4 + ..... + 32.33

=> 3S = 1.2.3 - 1.2.3 + 2.3.4 - 2.3.4 + ...... + 32.33.34

=> 3S = 32.33.34

=> S = \(\frac{32.33.34}{3}=11968\)

cau hỏi tương tự ko có mà!!!!!!!!!!!!!!!!!!!!!!!!!!!!

3C=1.2.3+2.3.(4-1)+3.4.(5-2)+...+2014.2015.(2016-2013)

3C=2014.2015.2016

C=2014.2015.2016:3