Ta có: 

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

<=> \(2n:\left(\frac{2}{2}+\frac{2}{3.2}+\frac{2}{4.3}+...+\frac{2}{\left(n+1\right).n}\right)=2020\)

<=> \(n:\left(1+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{n\left(n+1\right)}\right)=2020\)

<=> \(n:\left(1+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{n}-\frac{1}{n+1}\right)=2020\)

<=> \(n:\left(1-\frac{1}{n+1}\right)=2020\)

<=> \(n:\frac{n}{n+1}=2020\)

<=> n + 1 = 2020

<=> n = 2019

a/ \(=lim\frac{\left(-\frac{2}{3}\right)^n+1}{-2.\left(-\frac{2}{3}\right)^n+3}=\frac{1}{3}\)

b/ \(=lim\frac{\left(2-\frac{1}{n}\right)\left(1+\frac{1}{n}\right)\left(3+\frac{4}{n}\right)}{\left(\frac{5}{n}-6\right)^3}=\frac{2.1.3}{\left(-6\right)^3}=-\frac{1}{36}\)

c/ \(=lim\frac{5n+3}{\sqrt{n^2+5n+1}+\sqrt{n^2-2}}=\frac{5+\frac{3}{n}}{\sqrt{1+\frac{5}{n}+\frac{1}{n^2}}+\sqrt{1-\frac{2}{n}}}=\frac{5}{1+1}=\frac{5}{2}\)

d/ \(=lim\frac{5.\left(\frac{1}{2}\right)^n-6}{4.\left(\frac{1}{3}\right)^n+1}=\frac{-6}{1}=-6\)

e/ \(=-n^3\left(2+\frac{3}{n}-\frac{5}{n^2}+\frac{2020}{n^3}\right)=-\infty.2=-\infty\)

\(\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)...\left(1-\frac{1}{n+1}\right)\)

\(=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}...\frac{n}{n+1}\)

\(=\frac{1}{n+1}\)

\(1+\frac{1}{2}.\left(1+2\right)+\frac{1}{3}.\left(1+2+3\right)...+\frac{1}{20}.\left(1+2+3+...+20\right)\)

\(=1+\frac{1}{2}.2.3:2+\frac{1}{3}.3.4:2+\frac{1}{4}.4.5:2+...+\frac{1}{20}.20.21:2\)

\(=\frac{2}{2}+\frac{3}{2}+\frac{4}{2}+\frac{5}{2}+...+\frac{21}{2}\)

\(=\frac{2+3+4+5+...+21}{2}=115\)

\(a)\) Đặt \(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2010^2}\) ta có : 

\(A< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2009.2010}\)

\(A< \frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2009}-\frac{1}{2010}\)

\(A< 1-\frac{1}{2010}=\frac{2009}{2010}< 1\)

\(\Rightarrow\)\(A< 1\) ( đpcm ) 

Vậy \(A< 1\)

Chúc bạn học tốt ~ 

Ta có

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

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

Áp dụng vào bài toán ta có

\(\left(1-\frac{1}{1+2}\right)\left(1-\frac{1}{1+2+3}\right)...\left(1-\frac{1}{1+2+...+x}\right)=\frac{672}{2017}\)

\(=\frac{1.4}{2.3}.\frac{2.5}{3.4}...\frac{\left(x-1\right)\left(x+2\right)}{x\left(x+1\right)}=\frac{672}{2017}\)

\(\Leftrightarrow\frac{1}{3}.\frac{x+2}{x}=\frac{672}{2017}\)

\(\Leftrightarrow2017x+4034=2016x\)

1. \(\Leftrightarrow x=-4034\)

Đề đúng không thế sao t ra đáp số là số âm ta