\(A=\left(1+\frac{1}{1\cdot3}\right)\left(1+\frac{1}{2\cdot4}\right)\left(1+\frac{1}{3\cdot5}\right)\left(1+\frac{1}{4\cdot6}\right)...\left(1+\frac{1}{2020\cdot2022}\right)\) 

\(A=\frac{4}{1\cdot3}\cdot\frac{9}{2\cdot4}\cdot\frac{16}{3\cdot5}\cdot\frac{25}{4\cdot6}\cdot....\cdot\frac{4084441}{2021\cdot2022}\)

\(A=\frac{2\cdot2}{1\cdot3}\cdot\frac{3\cdot3}{2\cdot4}\cdot\frac{4\cdot4}{3\cdot5}\cdot\frac{5\cdot5}{4\cdot6}\cdot...\cdot\frac{2021\cdot2021}{2020\cdot2022}\)

\(A\frac{2\cdot2\cdot3\cdot3\cdot4\cdot4\cdot5\cdot5\cdot...\cdot2021\cdot2021}{1\cdot3\cdot2\cdot4\cdot3\cdot5\cdot4\cdot6\cdot....\cdot2020\cdot2022}\)

\(A=\frac{\left(2\cdot3\cdot4\cdot5\cdot...\cdot2021\right)\left(2\cdot3\cdot4\cdot5\cdot...\cdot2021\right)}{\left(1\cdot2\cdot3\cdot4\cdot...\cdot2020\right)\left(3\cdot4\cdot5\cdot6\cdot...\cdot2022\right)}\)

\(A=\frac{2021\cdot2}{2022}\)

\(A=\frac{4042}{2022}\)

\(A=\frac{2021}{1011}\left(đpcm\right)\)

Có số phần tử là

(90 - 2) : 4 + 1 = 23(phần tử)

HT

sai r bn ơi các số cách nhau lần lượt các số chăn từ lớn đến bé mà bn

\(\Leftrightarrow3\left(x-2\right)=21-18\)

\(\Leftrightarrow3\left(x-2\right)=3\)

\(\Leftrightarrow x-2=1\)

\(\Leftrightarrow x=3\)

21 - 3 . ( x - 2 ) = 18

       3 . ( x - 2 ) = 21 - 18 

       3 . ( x - 2 ) = 3

               x - 2  = 3 : 3

               x - 2 = 1

                   x  = 1 + 2

                   x  = 3

a) \(S=1.2+2.3+3.4+...+n\left(n+1\right)\)

\(3S=1.2.3+2.3.\left(4-1\right)+3.4.\left(5-2\right)+...+n\left(n+1\right)\left[\left(n+2\right)-\left(n-1\right)\right]\)

\(=1.2.3+2.3.4-1.2.3+...+n\left(n+1\right)\left(n+2\right)-\left(n-1\right)n\left(n+1\right)\)

\(=n\left(n+1\right)\left(n+2\right)\)

\(\Rightarrow S=\frac{n\left(n+1\right)\left(n+2\right)}{3}\)

b) \(S=1.2.3+2.3.4+...+n\left(n+1\right)\left(n+2\right)\)

\(4S=1.2.3.4+2.3.4.\left(5-1\right)+...+n\left(n+1\right)\left(n+2\right)\left[\left(n+3\right)-\left(n-1\right)\right]\)

\(=1.2.3.4+2.3.4.5-1.2.3.4+...+n\left(n+1\right)\left(n+2\right)\left(n+3\right)-\left(n-1\right)n\left(n+1\right)\left(n+2\right)\)

\(=n\left(n+1\right)\left(n+2\right)\left(n+2\right)\)

\(S=\frac{n\left(n+1\right)\left(n+2\right)\left(n+3\right)}{4}\)

c) \(S=1.4+2.5+3.6+...+n\left(n+3\right)\)

\(=1.2+1.2+2.3+2.2+3.4+3.2+...+n\left(n+1\right)+2n\)

\(=\left(1.2+2.3+3.4+...+n\left(n+1\right)\right)+2\left(1+2+3+...+n\right)\)

\(=\frac{n\left(n+1\right)\left(n+2\right)}{3}+n\left(n+1\right)\)

\(=\frac{n\left(n+1\right)\left(n+5\right)}{3}\)

\(\frac{-2}{5}\) \(.\) \(\frac{1}{8}\) =  \(\frac{1}{8}\) \(.\) \(\frac{3}{5}\)=   \(\frac{3}{40}\)

 - Hok t - 

Trả lời :

\(-\frac{2}{5}\cdot\frac{1}{8}=\frac{1}{8}\cdot\frac{3}{5}=\frac{3}{40}\)

1k đê

~HT~

có 13-x=15

x=13-15

x=-2

Vậy B={-2}

x+1/3=3/5

x=4/15

HT

x + 1/3 = 3/5

       x = 3/5 - 1/3

       x = 4/15

hoc tot