câu b bài 2:

\(\dfrac{1^2}{1\cdot2}\cdot\dfrac{2^2}{2\cdot3}\cdot\dfrac{3^2}{3\cdot4}\cdot\dfrac{4^2}{4\cdot5}\)

\(=\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot\dfrac{3}{4}\cdot\dfrac{4}{5}\)

\(=\dfrac{1}{5}\)

câu a bài 2:

\(\dfrac{1}{1\cdot2\cdot3}+\dfrac{1}{2\cdot3\cdot4}+\dfrac{1}{3\cdot4\cdot5}+...+\dfrac{1}{10\cdot11\cdot12}\)

\(=\dfrac{1}{1}-\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{4}-...-\dfrac{1}{12}\)

\(=1-\dfrac{1}{12}=\dfrac{11}{12}\)

\(A=\left(6-\dfrac{2}{3}+\dfrac{1}{2}\right)-\left(5+\dfrac{5}{3}-\dfrac{3}{2}\right)-\left(3-\dfrac{7}{3}+\dfrac{5}{2}\right)\\ A=\dfrac{35}{6}-\dfrac{31}{6}-\dfrac{19}{6}=-\dfrac{15}{6}\)

\(a:=\left(\dfrac{1}{4}+\dfrac{3}{4}\right)+\left(\dfrac{1}{5}+\dfrac{2}{5}\right)\\ =1+\dfrac{3}{5}=1\dfrac{3}{5}\\ b:=\left(\dfrac{2}{7}+\dfrac{1}{7}\right)+\left(\dfrac{5}{14}+\dfrac{3}{14}\right)\\ =\dfrac{3}{7}+\dfrac{8}{14}\\ =\dfrac{3}{7}+\dfrac{4}{7}=1\)

a) = 1/2 - 1/2 + 1/3 -1/3 + 1/4 - 1/4 + 1/5 - 1/5 + 1/6

    =  0 + 0 + 0 + 0 + 1/6

    = 1/6

b) 2/3 + 2/4 - 2/4 + 2/5 - 2/5 + 2/6 - 2/6 + 2/7 - 2/7 + 28

=  2/3 + 28

=     86/3

[tick cho mik nha]

a) \(1+2+3+...+2021=\dfrac{\left(2021+1\right)\left(\dfrac{2021-1}{1}+1\right)}{2}=2043231\)

b) Giống a

c) \(11+13+15+...131=\dfrac{\left(131+11\right)\left(\dfrac{131-11}{2}+1\right)}{2}=4331\)

a. 1 + 2 + 3 + ... + 2021

= 1 + 2 + 3 + ... + 2020) + 2021

Ta có số cặp từ 1 đên 2020 là: 2020 : 2 = 1010(cặp)

Ta có tổng số đầu và số cuối từ 1 đến 2020 là: 2020 + 1 = 2021

=> Tổng là: 2021 . 1010 + 2021 = 2043231

`a)(1-1/2)xx(1-1/3)xx(1-1/4)xx(1-1/5)`

`=1/2xx2/3xx3/4xx4/5`

`=[1xx2xx3xx4]/[2xx3xx4xx5]`

`=1/5`

`b)(1-3/4)xx(1-3/7)xx(1-3/10)xx(1-3/13)xx .... xx(1-3/97)xx(1-3/100)`

`=1/4xx4/7xx7/10xx10/13xx .... xx94/97xx97/100`

`=[1xx4xx7xx10xx...xx94xx97]/[4xx7xx10xx13xx....xx97xx100]`

`=1/100`

A=1+1/2+1/3+1/6+(1/12+1/15+1/20+1/30)+1/35

=71/35+7/30=95/42

\(a,=\dfrac{1}{3}\times\left(\dfrac{1}{2}+\dfrac{1}{3}\right)=\dfrac{1}{3}\times\dfrac{5}{6}=\dfrac{5}{18}\\ b,=\dfrac{4}{5}\times\left(\dfrac{1}{2}-\dfrac{1}{3}\right)=\dfrac{4}{5}\times\dfrac{1}{6}=\dfrac{2}{15}\\ c,=456\times99-6\times99+456\\ =456\times\left(99+1\right)-594\\ =456\times100-594=45600-594=45006\\ d,=101\times\left(101-1\right)=101\times100=10100\)

\(a,\left(1-\dfrac{1}{10}\right)\times\left(1-\dfrac{1}{11}\right)\times\left(1-\dfrac{1}{12}\right)\times\left(1-\dfrac{1}{13}\right)\times\left(1-\dfrac{1}{14}\right)\times\left(1-\dfrac{1}{15}\right)=\dfrac{9}{10}\times\dfrac{10}{11}\times\dfrac{11}{12}\times\dfrac{12}{13}\times\dfrac{13}{14}\times\dfrac{14}{15}=\dfrac{9}{15}=\dfrac{3}{5}\)

\(b,\dfrac{2013\times2012-2}{2011+2011\times2013}=\dfrac{\left(2014-1\right)\times2012-2}{2011\times\left(2013+1\right)}=\dfrac{2014\times2012-2012-2}{2011\times2014}=\dfrac{2014\times2012-2014}{2011\times2014}=\dfrac{2014\times\left(2012-1\right)}{2011\times2014}=\dfrac{2011\times2014}{2011\times2014}=1\)

3/5,1

1. \(\left(\dfrac{1}{99}+\dfrac{12}{999}-\dfrac{123}{9999}\right).\left(\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{6}\right)\)

\(=\left(\dfrac{1}{99}+\dfrac{12}{999}-\dfrac{123}{9999}\right).0\)

\(=0\)

Bài 1: 

Ta có: \(A=\left(\dfrac{1}{99}+\dfrac{12}{999}-\dfrac{123}{9999}\right)\cdot\left(\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{6}\right)\)

\(=\left(\dfrac{1}{99}+\dfrac{12}{999}-\dfrac{123}{9999}\right)\cdot\left(\dfrac{3}{6}-\dfrac{2}{6}-\dfrac{1}{6}\right)\)

=0