A = \(\dfrac{1}{1+2}\) + \(\dfrac{1}{1+2+3}\) + \(\dfrac{1}{1+2+3+4}\)+...+ \(\dfrac{1}{1+2+3+...+2020}\)

Ta có S = 1 + 2 + ...+ n 

Dãy số trên là dãy số cách đều với khoảng cách là: 2 - 1 = 1

Số số hạng của dãy số trên là: (n-1): 1 + 1 = n

Áp dụng công thức tính tổng của dãy số cách đều ta có tổng trên là:

S = (n+1)\(\times\) n : 2

Áp dụng công thức tính tổng S trên vào biểu thức A ta có:

A = \(\dfrac{1}{\left(2+1\right)\times2:2}\)+\(\dfrac{1}{\left(3+1\right)\times3:2}\)+...+\(\dfrac{1}{\left(2020+1\right)\times2020:2}\)

A =  \(\dfrac{1}{2\times3:2}\)  + \(\dfrac{1}{3\times4:2}\)+ \(\dfrac{1}{4\times5:2}\)+...+\(\dfrac{1}{2020\times2021:2}\)

A = \(\dfrac{2}{2\times3}\) + \(\dfrac{2}{3\times4}\) + \(\dfrac{2}{4\times5}\)+...+ \(\dfrac{2}{2020\times2021}\)

A = \(2\) \(\times\)( \(\dfrac{1}{2\times3}\) + \(\dfrac{1}{3\times4}\)+ \(\dfrac{1}{4\times5}\)+...+ \(\dfrac{1}{2020\times2021}\))

A = 2 \(\times\)( \(\dfrac{1}{2}\) - \(\dfrac{1}{3}\) + \(\dfrac{1}{3}\) - \(\dfrac{1}{4}\)+\(\dfrac{1}{4}\) - \(\dfrac{1}{5}\)+...+ \(\dfrac{1}{2020}\)- \(\dfrac{1}{2021}\))

A = 2\(\times\)( \(\dfrac{1}{2}\) - \(\dfrac{1}{2021}\))

A = 1 - \(\dfrac{2}{2021}\)

A = \(\dfrac{2021-2}{2021}\)

A = \(\dfrac{2019}{2021}\)

thật sự mị ko biết

B=1,59(285714)

HOK TỐT

NHỚ LÀ NGƯỜI TA GIÚP MÌNH LÀ PHẢI CẢM ƠN NGƯỜI TA ĐÓ NHA.

HOK TỐT

help me !!!!!!!

\(=2021\cdot2\cdot\left(1+\dfrac{1}{2}:\dfrac{3}{2}-\dfrac{4}{3}\right)=4042\cdot\left(1+\dfrac{1}{3}-\dfrac{4}{3}\right)=0\)

M = 0

\(M=\left(\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{2020^2}\right)\left(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2020^2}\right)-\left(\frac{1}{1^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2020^2}\right)\left(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2020^2}\right)\)

\(M=\left(\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{2020^2}\right)\left(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2020^2}\right)(1-1)\)

\(M=\left(\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{2020^2}\right)\left(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2020^2}\right).0\)

\(M=0\)

Vì số bị trừ và số trừ gồm hai tích đảo ngược nhau nên M=0

= 1.495676488x10^-3

\(A=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)...\left(1-\frac{1}{2021}\right)\)

\(A=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.....\frac{2020}{2021}\)

\(A=\frac{1}{2021}\)

Ta có \(B=\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).\left(1-\frac{1}{4}\right)...\left(1-\frac{1}{2020}\right)\)

\(=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}...\frac{2019}{2020}=\frac{1.2.3...2019}{2.3.4...2020}=\frac{1}{2020}\)

Lại có : \(A=\left(1\frac{1}{2}\right).\left(1\frac{1}{3}\right).\left(1\frac{1}{4}\right)...\left(1\frac{1}{2020}\right)\)

\(=\frac{3}{2}.\frac{4}{3}.\frac{5}{4}...\frac{2021}{2020}=\frac{3.4.5...2021}{2.3.4...2020}=\frac{2021}{2}\)

Khi đó \(\frac{A}{B}=\frac{\frac{2021}{2}}{\frac{1}{2020}}=\frac{2021}{2}.2020=2041210\)

Ta có :

B = \(\dfrac{1}{2020}+\dfrac{2}{2019}+\dfrac{3}{2018}+...+\dfrac{2019}{2}+\dfrac{2020}{1}\)

B = \(\left(\dfrac{1}{2020}+1\right)+\left(\dfrac{2}{2019}+1\right)+\left(\dfrac{3}{2018}+1\right)+...+\left(\dfrac{2019}{2}+1\right)+1\)

B = \(\dfrac{2021}{2020}+\dfrac{2021}{2019}+\dfrac{2021}{2018}+...+\dfrac{2021}{2}+1\)

B = \(2021\left(\dfrac{1}{2021}+\dfrac{1}{2020}+\dfrac{1}{2019}+...+\dfrac{1}{2}\right)\)  (1)

Mà A = \(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2021}\)   (2)

Từ (1) và (2) \(\Rightarrow\) \(\dfrac{A}{B}=\dfrac{1}{2021}\)

Ta có: \(B=\dfrac{1}{2020}+\dfrac{2}{2019}+\dfrac{3}{2018}+...+\dfrac{2019}{2}+\dfrac{2020}{1}\)

\(=\left(\dfrac{1}{2020}+1\right)+\left(\dfrac{2}{2019}+1\right)+\left(\dfrac{3}{2018}+1\right)+...+\left(\dfrac{2019}{2}+1\right)+1\)

\(=\dfrac{2021}{2020}+\dfrac{2021}{2019}+\dfrac{2021}{2018}+...+\dfrac{2021}{2}+\dfrac{2021}{2021}\)

Suy ra: \(\dfrac{A}{B}=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2021}}{2021\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2021}\right)}=\dfrac{1}{2021}\)

Giải:

Ta có:

\(B=\dfrac{1}{2020}+\dfrac{2}{2019}+\dfrac{3}{2018}+...+\dfrac{2019}{2}+\dfrac{2020}{1}\) 

\(B=1+\left(\dfrac{1}{2020}+1\right)+\left(\dfrac{2}{2019}+1\right)+\left(\dfrac{3}{2018}+1\right)+...+\left(\dfrac{2019}{2}+1\right)\) 

\(B=\dfrac{2021}{2021}+\dfrac{2021}{2020}+\dfrac{2021}{2019}+\dfrac{2021}{2018}+...+\dfrac{2021}{2}\) 

\(B=2021.\left(\dfrac{1}{2021}+\dfrac{1}{2020}+\dfrac{1}{2019}+\dfrac{1}{2018}+...+\dfrac{1}{2}\right)\) 

\(\Rightarrow\dfrac{A}{B}=\dfrac{\left[2021.\left(\dfrac{1}{2021}+\dfrac{1}{2020}+\dfrac{1}{2019}+\dfrac{1}{2018}+...+\dfrac{1}{2}\right)\right]}{\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2021}\right)}=2021\) 

Vậy \(\dfrac{A}{B}=2021\)