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10 tháng 3 2020

ta có \(A=\frac{2020^{10}+2}{2020^{11}+2}=>2020A=\frac{2020^{11}+4040}{2020^{11}+2}=1+\frac{4038}{2020^{11}+2}\)(1)

\(B=\frac{2020^{11}+2}{2020^{12}+2}=>2020B=\frac{2020^{12}+4040}{2020^{12}+2}=1+\frac{4038}{2012^{12}+2}\)(2)

từ 1 and 2 => 2020B<2020A

=> A>B

Ta có B=\(\frac{2020^{11}+2}{2020^{12}+2}\)

suy ra \(B< \frac{\left(2020^{11}+2\right)+2018}{\left(2020^{12}+2\right)+2018}=\frac{2020^{11}+2020}{2020^{12}+2020}=\frac{2020\left(2020^{10}+2\right)}{2020\left(2020^{11}+2\right)}=\frac{2020^{10}+2}{2020^{11}+2}\)

nên A > B

A=2020^10+2/2020^11+2

⇒ 2020A=2020^11+2.2020/2020^11+2

= 1+2.2020−2/2020^11+2

B=2020^11+2/2020^12+2

⇒ 2020B=2020^12+2.2020/2020^12+2

= 1+2.2020−2/2020^12+2

Vì 2020^12+2>2020^11+2

⇒ 2.2020−2/2020^11+2<2.2020−2/2020^12+2

⇒ 2020A<2020B

⇒ A<B

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

30 tháng 7 2020

Ta có : A = \(\frac{10^{2020}+1}{10^{2021}+1}\)

=> 10A = \(\frac{10^{2021}+10}{10^{2021}+1}=1+\frac{9}{10^{2021}+1}\)

Lại có : \(B=\frac{10^{2021}+1}{10^{2022}+1}\)

=> \(10B=\frac{10^{2022}+10}{10^{2022}+1}=1+\frac{9}{10^{2022}+1}\)

Vì \(\frac{9}{10^{2022}+1}< \frac{9}{10^{2021}+1}\)

=> \(1+\frac{9}{10^{2022}+1}< 1+\frac{9}{10^{2022}+1}\)

=> 10B < 10A

=> B < A

b) Ta có : \(\frac{2019}{2020+2021}< \frac{2019}{2020}\)

Lại có : \(\frac{2020}{2020+2021}< \frac{2020}{2021}\)

=> \(\frac{2019}{2020+2021}+\frac{2020}{2020+2021}< \frac{2019}{2020}+\frac{2020}{2021}\)

=> \(\frac{2019+2020}{2020+2021}< \frac{2019}{2020}+\frac{2020}{2021}\)

=> B < A

13 tháng 2 2022

sai rồi

14 tháng 5 2019

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

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

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

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

HQ
Hà Quang Minh
Giáo viên
7 tháng 10 2023

a) Ta có: \( - 2 = \frac{{ - 2}}{1} = \frac{{ - 40}}{{20}}\)

\(\frac{{ - 11}}{5} = \frac{{ - 44}}{{20}} < \frac{{ - 40}}{{20}}\) nên \(\frac{{ - 11}}{5} < -2\).

\(\frac{{ - 7}}{4} = \frac{{ - 7.5}}{{4.5}} = \frac{{ - 35}}{{20}} > \frac{{ - 40}}{{20}}\) nên \(\frac{{ - 7}}{4} > -2\)

Vậy \(\frac{{ - 11}}{5} < \frac{{ - 7}}{4}\).

b) Ta có: \(\frac{{2020}}{{ - 2021}} = \frac{{ - 2020}}{{2021}} > \frac{{ - 2022}}{{2021}}\)

Vậy \(\frac{{2020}}{{ - 2021}} > \frac{{ - 2022}}{{2021}}\)

29 tháng 4 2020

Ta có: 

\(10A=\frac{10^{2015}+20200}{10^{2015}+2020}=1+\frac{18180}{10^{2015}+2020}\)

\(10B=\frac{10^{2016}+20200}{10^{2016}+2020}=1+\frac{18180}{10^{2016}+2020}\)

Vì \(10^{2016}+2020>2^{2015}+2020\)

=> \(\frac{18180}{10^{2016}+2020}< \frac{18180}{10^{2015}+2020}\)

=> \(1+\frac{18180}{10^{2016}+2020}< 1+\frac{18180}{10^{2015}+2020}\)

=> 10B < 10A

=> B<A

29 tháng 4 2020

\(A=\frac{10^{2014}+2020}{10^{2015}+2020}\)\(< \) \(B=\frac{10^{2015}+2020}{10^{2016}+2020}\)

chúc bạn học tốt

study well

22 tháng 5 2019

đặt 22018 = a ; 32019 = b ; 52020 = c

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

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

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

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

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

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

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

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

22 tháng 5 2019

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

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

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

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

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

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