ta có :\(E=\frac{2019^{2019}+1}{2019^{2020}+1}\Leftrightarrow2019\cdot E=\frac{2019^{2020}+2019}{2019^{2020}+1}=1+\frac{2019}{2019^{2020}+1}\)

\(F=\frac{2019^{2020}+1}{2019^{2021}+1}\Leftrightarrow2019\cdot F=\frac{2019^{2021}+2019}{2019^{2021}+1}=1+\frac{2019}{2019^{2021}+1}\)

vì \(\frac{2019}{2019^{2020}+1}>\frac{2019}{2019^{2021}+1}\) nên E>F

E=2019 x 2019 x 2019 x ........ x 2019 x2019 +1 /2019 x 2019 x 2019 x.........x 2019 x 2019 + 1

E=1+1/2019+1

E=2/2020

E=1/1010

F=2019 x 2019 x 2019 x .......... x 2019 x 2019 +1 / 2019 x 2019 x 2019 x ....... x 2019 x 2019 +1

F= 1+1/2019+1

F=2/2020

F=1/1010

từ đó ta có E=F(=1/1010)

nghỉ sao rút gọn được vậy có dấu + mà(ví dụ\(\frac{2\cdot2+1}{2\cdot2\cdot2+1}=\frac{5}{9}\ne\frac{2}{3}\))

Tham khảo

https://hoc24.vn/hoi-dap/question/814814.html

B=11.2+13.4+15.6+....+12019.2020

⇒2B=21.2+23.4+25.6+....+22019.2020

<1+12.3+13.4+14.5+15.6+....+12018.2019+12019.2020

2B<1+3−22.3+4−33.4+5−44.5+....+2019−20182018.2019+2020−20192019.2020

2B<1+12−13+13−14+...+12019−12020

2B<1+12−12020<1+12

B<34

---------------------

Đặt 22018=a;32019=b;52020=c(a,b,c>0)

A=aa+b+bb+c+cc+a>aa+b+c+ba+b+c+ca+b+c=1

⇒A>1>34>B

2019 + 2019 : 0,5 + 2019 : 0,2 + 2019 : 0,125

= 2019 x 1 + 2019 x 2 + 2019 x 5 + 2019 x 8

= 2019 x ( 1 + 2 + 5 + 8 )

= 2019 x 16

= 32304

Trl :

2019 + 2019 : 0,5 + 2019 : 0,2 + 2019 : 0,125

= 2019 x 1 + 2019 x 2 + 2019 x 5 + 2019 x 8 

= 2019 x ( 1 + 2 + 5 + 8 )

= 2019 x 16

= 32304

=(2015/ 2019 + 3/2019 + 1/2019 ) : 1/2

= 2019/2019 x 2

= 1 x2 

=2

2015/2019:1/2+3/2019:1/2+1/2019:1/2

=(2015/2019+3/2019+1/2019):1/2

=1:1/2

=2

k cho mink nha

2015/2019 : 1/2 + 3/2019 : 1/2 + 1/2019 : 1/2

= ( 2015/2019 + 3/2019 + 1/2019 ) : 1/2

= 1 : 1/2

= 2

Chúc bn hok tốt ! ^_^

Đặt D = \(\dfrac{1}{2}\) + \(\dfrac{1}{2^2}\) + \(\dfrac{1}{2^3}\) + ...... + \(\dfrac{1}{2^{2019}}\) 

      ⇔ 2D = 1 + \(\dfrac{1}{2}\) + \(\dfrac{1}{2^2}\) + ...... + \(\dfrac{1}{2^{2018}}\)  

      ⇔  D   = 1 - \(\dfrac{1}{2^{2019}}\) 

     ⇒ A  = (1 - \(\dfrac{1}{2^{2019}}\)) : (1 - \(\dfrac{1}{2^{2019}}\))

       ⇒ A = 1

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 \(N=2018\frac{1009}{2020}.\)