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

Đặt \(A=\frac{\frac{1}{2020}+\frac{2}{2019}+\frac{3}{2018}+...+\frac{2019}{2}+\frac{2020}{1}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2021}}\)

\(A=\frac{1+\left(\frac{1}{2020}+1\right)+\left(\frac{2}{2019}+1\right)+\left(\frac{3}{2018}+1\right)+...+\left(\frac{2019}{2}+1\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2021}}\)

\(A=\frac{\frac{2021}{2021}+\frac{2021}{2020}+\frac{2021}{2019}+...+\frac{2021}{2}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2021}}\)

\(A=\frac{2021\left(\frac{1}{2021}+\frac{1}{2020}+\frac{1}{2019}+...+\frac{1}{2}\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2021}}=2021\)

Đặt \(a=\frac{1}{1^2}+\frac{1}{2^2}+...+\frac{1}{2019^2}\)

\(b=\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{2020^2}\)

Khi đó : \(D=ab-\left(b+1\right)\left(a-1\right)\)

\(\Rightarrow D=ab-\left(ab+a-b-1\right)\)

\(\Rightarrow D=b-a+1=\frac{1}{2020^2}-1+1=\frac{1}{2020^2}\)

Sửa đề \(\frac{2019}{1}+\frac{2018}{2}+...+\frac{1}{2019}\)

Ta có: \(\frac{2019}{1}+\frac{2018}{2}+...+\frac{1}{2019}\)

\(=\left(2019+1\right)+\left(\frac{2018}{2}+1\right)+...+\left(\frac{1}{2019}+1\right)-2019\)

\(=2020+\frac{2020}{2}+...+\frac{2020}{2019}+\frac{2020}{2020}-2020\)

\(=\frac{2020}{2}+...+\frac{2020}{2019}+\frac{2020}{2020}\)

\(=2020.\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2020}\right)\)Thay vào biểu thức A ta được:

\(A=\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2020}}{2020.\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2020}\right)}=\frac{1}{2020}\)

Ta có :\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=6\Rightarrow\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2=36\Rightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right)=36\)

 \(\Rightarrow\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}=12\)

\(\Rightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}=\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\)

\(\Rightarrow\frac{2}{a^2}+\frac{2}{b^2}+\frac{2}{c^2}=\frac{2}{ab}+\frac{2}{bc}+\frac{2}{ca}\)

=> \(\frac{2}{a^2}+\frac{2}{b^2}+\frac{2}{c^2}-\frac{2}{ab}-\frac{2}{bc}-\frac{2}{ca}=0\)

=> \(\left(\frac{1}{a^2}-\frac{2}{ab}+\frac{1}{b^2}\right)+\left(\frac{1}{b^2}-\frac{2}{bc}+\frac{1}{c^2}\right)+\left(\frac{1}{c^2}-\frac{2}{ac}+\frac{1}{a^2}\right)=0\)

=> \(\left(\frac{1}{a}-\frac{1}{b}\right)^2+\left(\frac{1}{b}-\frac{1}{c}\right)^2+\left(\frac{1}{c}-\frac{1}{a}\right)^2=0\)

=> \(\hept{\begin{cases}\frac{1}{a}-\frac{1}{b}=0\\\frac{1}{b}-\frac{1}{c}=0\\\frac{1}{c}-\frac{1}{a}=0\end{cases}}\Rightarrow\frac{1}{a}=\frac{1}{b}=\frac{1}{c}\)

Khi đó \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=6\Leftrightarrow3\frac{1}{a}=6\Rightarrow\frac{1}{a}=2\Leftrightarrow\frac{1}{a}=\frac{1}{b}=\frac{1}{c}=2\)

Khi đó  Đặt P = \(\left(\frac{1}{a}-3\right)^{2020}+\left(\frac{1}{b}-3\right)^{2020}+\left(\frac{1}{c}-3\right)^{2020}\)

= (2 - 3)²⁰²⁰ + (2 - 3)²⁰²⁰ + (2 - 3)²⁰²⁰

= 1 + 1 + 1 = 3

Vậy P = 3 

1) Ta có: \(2020^2=\left(2019+1\right)^2=2019^2+2.2019+1.\)

\(\Rightarrow1+2019^2=2020^2-2.2019\)

\(\Rightarrow M=\sqrt{1+2019^2+\frac{2019^2}{2020^2}}+\frac{2019}{2020}=\sqrt{2020^2-2.2019+\frac{2019^2}{2020^2}}+\frac{2019}{2020}\)

\(=\sqrt{2020^2-2.2020.\frac{2019}{2020}+\left(\frac{2019}{2020}\right)^2}+\frac{2019}{2020}\)

\(=\sqrt{\left(2020-\frac{2019}{2020}\right)^2}+\frac{2019}{2020}=2020-\frac{2019}{2020}+\frac{2019}{2020}\)

\(=2020\)

Vậy M=2020.

2) Xét  : \(k\in N;k\ge2\)ta có:

\(\left(1+\frac{1}{k-1}-\frac{1}{k}\right)^2=1+\frac{1}{\left(k-1\right)^2}+\frac{1}{k^2}+\frac{2}{k-1}-\frac{2}{\left(k-1\right)k}-\frac{2}{k}\)

                                          \(=1+\frac{1}{\left(k-1\right)^2}+\frac{1}{k^2}+\frac{2}{k-1}-\frac{2}{k-1}+\frac{2}{k}-\frac{2}{k}\)

\(\Rightarrow\left(1+\frac{1}{k-1}-\frac{1}{k}\right)^2=1+\frac{1}{\left(k-1\right)^2}+\frac{1}{k^2}\)

\(\Rightarrow\sqrt{1+\frac{1}{\left(k-1\right)^2}+\frac{1}{k^2}}=1+\frac{1}{k-1}+\frac{1}{k}\)

Cho \(k=3,4,...,2020.\)Ta có:

\(N=\sqrt{1+\frac{1}{2^2}+\frac{1}{3^2}}+\sqrt{1+\frac{1}{3^2}+\frac{1}{4^2}}+...+\sqrt{1+\frac{1}{2019^2}+\frac{1}{2020^2}}\)

\(=\left(1+\frac{1}{2}-\frac{1}{3}\right)+\left(1+\frac{1}{3}-\frac{1}{4}\right)+...+\left(1+\frac{1}{2018}-\frac{1}{2019}\right)+\left(1+\frac{1}{2019}-\frac{1}{2020}\right)\)

\(=2018+\frac{1}{2}-\frac{1}{2020}=2018\frac{1009}{2020}\)

Vậy \(N=2018\frac{1009}{2020}.\)

HD: Vũ Phương Vy  em chỉ cần đặt ts c rồi rút gọn

ko chép lại đề nha

=\(A=\frac{2\left(1-\frac{2}{19}+\frac{2}{23}\right)-\frac{1}{1010}}{3\left(1-\frac{1}{19}+\frac{1}{23}-\frac{1}{2020}\right)}\)\(.\frac{4\left(1-\frac{1}{29}+\frac{1}{41}\right)-\frac{1}{505}}{5\left(1-\frac{1}{29}+\frac{1}{41}\right)-\frac{1}{404}}\)

rồi em chỉ cần rút gọn tiếp 

p/s đến đây thấy đề kì kì sao đó 

em chek lại đề đc k

đề đúng rồi đó ạ

đề đúng mà ko sai đâu nếu bạn ko tin thì xem bài 5 của đề này nhé : https://online-learning-izteach-aws-source-bucket.s3-ap-southeast-1.amazonaws.com/resource/49714800-4b02-11ea-a775-01bd84d6799a/4baf4091-c5ae-4d0f-819c-e9ee9eedd6a8/de-so-13-on-tap-va-cung-co-toan-6-pdf

\(F=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{2019\cdot2020}\)

Trả lời:

\(F=1.\frac{1}{2}+\frac{1}{2}.\frac{1}{3}+\frac{1}{3}.\frac{1}{4}+...+\frac{1}{2019}.\frac{1}{2020}\)

\(F=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2019.2020}\)

\(F=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2019}-\frac{1}{2020}\)

\(F=\frac{1}{1}-\frac{1}{2020}\)

\(F=\frac{2019}{2020}\)

\(F=1.\frac{1}{2}+\frac{1}{2}.\frac{1}{3}+\frac{1}{3}.\frac{1}{4}+...+\frac{1}{2019}.\frac{1}{2020}\)

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

\(=1-\frac{1}{2020}=\frac{2020}{2020}-\frac{1}{2020}=\frac{2019}{2020}\)

Vậy \(F=\frac{2019}{2020}.\)

A = 2017 - ( 1 -\(\frac{3}{4}\)) - ( 1 - \(\frac{3}{5}\)) -...- ( 1 - \(\frac{3}{2020}\)) : Còn lại

A = 2017 - 1 + \(\frac{3}{4}\) -  1 + \(\frac{3}{5}\) -... - 1 +\(\frac{3}{2020}\) : Còn lại

A = 0 + \(\frac{3}{4}\)+\(\frac{3}{5}\)+\(\frac{3}{2020}\) : CÒn lại

A = 3 x ( \(\frac{1}{4}\)+\(\frac{1}{5}\)+...+ \(\frac{1}{2020}\)) : ( \(\frac{1}{4x5}\)+\(\frac{1}{5x5}\) + ... )

A = 3 : \(\frac{1}{5}\)

A = 15

= 15 nha

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bằng 15 mk chắc chắn vì mk thi violympic vòng 17 mk được 300 mà

bạn kết bạn với mk thì có bài nào khó thì mk sẽ đáp ứng