đặt 2²⁰¹⁸ = a ; 3²⁰¹⁹ = b ; 5²⁰²⁰ = c

Ta có : \(A=\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{a+c}>\frac{a}{a+b+c}+\frac{b}{a+b+c}+\frac{c}{a+b+c}=1\)

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

\(2B=\frac{2}{1.2}+\frac{2}{3.4}+...+\frac{2}{2019.2020}\)

\(< 1+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2018.2019}+\frac{1}{2019.2020}\)

\(2B< 1+\frac{3-2}{2.3}+\frac{4-3}{3.4}+....+\frac{2019-2018}{2018.2019}+\frac{2020-2019}{2019.2020}\)

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

\(B< \frac{3}{4}\)

\(\Rightarrow A>1>\frac{3}{4}>B\)

Mình chỉ biết cách tính B thôi, đây nhé:

B= \(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{2019.2020}\)

B=\(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{2019}-\frac{1}{2020}\)

\(B=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{2019}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{2020}\right)\)

\(B=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{2019}+\frac{1}{2020}\right)-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{2020}\right)\)

\(B=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{2019}+\frac{1}{2020}\right)-2\frac{1}{2}\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{1010}\right)\)

\(B=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{2019}+\frac{1}{2020}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{1010}\right)\)

\(B=\frac{1}{1011}+\frac{1}{1012}+....+\frac{1}{2019}+\frac{1}{2020}\)

\(\hept{\begin{cases}A=-\frac{1}{2020}-\frac{3}{2019^2}-\frac{5}{2019^3}-\frac{7}{2019^4}^{ }\\B=-\frac{1}{2020}-\frac{7}{2019^2}-\frac{5}{2019^3}-\frac{3}{2019^4}\end{cases}}\)

=>\(A-B=-\frac{1}{2020}-\frac{3}{2019^2}-\frac{5}{2019^3}-\frac{7}{2019^4}+\frac{1}{2020}+\frac{7}{2019^2}+\frac{5}{2019^3}+\frac{3}{2019^4}\)

\(=>A-B=\left(-\frac{3}{2019^2}+\frac{7}{2019^2}\right)+\left(-\frac{7}{2019^4}+\frac{3}{2019^4}\right)\)

=>\(A-B=\frac{4}{2019^2}+-\frac{4}{2019^4}\)

=>\(A-B=\frac{2019^2.4}{2019^4}-\frac{4}{2019^4}\)

=>\(A>B\)

cách này mình tự nghĩ 

thank you \(v\text{er}y^{1000000000000}\)much

B= 1/1.2+1/2.3+...+1/2019.2020

B=1/1-1/2+1/2-1/3+...+1/2019-1/2020

B=1-1/2020=2020/2020-1/2020=2019/2020

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

\(A=\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}>\frac{a}{a+b+c}+\frac{b}{a+b+c}+\frac{c}{a+b+c}=\frac{a+b+c}{a+b+c}=1.\) 

Với  :   \(a=2^{2018};.b=3^{2019};,c=5^{2020}.\) 

Và   :   \(B=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2019.2020}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2019}-\frac{1}{2020}\Leftrightarrow\) 

             \(B=1-\frac{1}{2020}< 1< A\)

XD: best tiếng anh chuyển sang toán ak!?

\(B1:\)

\(M=\left(1+\frac{7}{9}\right)\left(1+\frac{7}{20}\right)\left(1+\frac{7}{33}\right)...\left(1+\frac{7}{10800}\right)\)

\(=\frac{16}{9}\cdot\frac{27}{20}\cdot\frac{40}{33}\cdot\cdot\cdot\frac{10807}{10800}\)

\(=\frac{8.2}{9.1}\cdot\frac{9.3}{10.2}\cdot\frac{10.4}{11.3}\cdot\cdot\cdot\frac{57.51}{58.50}\)

\(=\frac{\left(8.9.10...57\right)\left(2.3.4...51\right)}{\left(9.10.11...58\right).\left(1.2.3...50\right)}\)

\(=\frac{8.51}{58.1}=\frac{204}{29}\)

Vậy.....

\(M=\left(1+\frac{7}{9}\right)\left(1+\frac{7}{20}\right)\left(1+\frac{7}{33}\right)...\left(1+\frac{7}{10800}\right)\)

\(M=\frac{16}{9}.\frac{27}{20}.\frac{40}{33}...\frac{10807}{10800}\)

\(M=\frac{8.2}{9.1}.\frac{9.3}{10.2}.\frac{10.4}{11.3}...\frac{107.101}{108.100}\)

\(M=\frac{\left(2.3.4...101\right)\left(8.9.10...107\right)}{\left(1.2.3...100\right)\left(9.10.11...108\right)}\)

\(M=\frac{101.8}{108}\)

\(M=\frac{202}{27}\)

k mình nha . câu 2 tí nữa mình gửi

câu 2 đau đầu thế

A<13 tick minh nha ban

Đáp án là A<13