So sánh A=\(\frac{2020^{10}+2}{2020^{11}+2}\)và B=\(\frac{2020^{11}+2}{2020^{12}+2}\)
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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
10 tháng 3 2020
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
8 tháng 3 2020
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
14 tháng 5 2019
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
XO
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
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 : \(B=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2019.2020}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2019}-\frac{1}{2020}\Leftrightarrow\)
\(B=1-\frac{1}{2020}< 1< A\)
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
26 tháng 8 2020
a) Ta có : \(\frac{-60}{12}=-5=-\frac{25}{5}\)
\(-0,8=-\frac{8}{10}=-\frac{4}{5}\)
Mà -25 < -4 nên \(\frac{-25}{5}< \frac{-4}{5}\)=> \(\frac{-60}{12}< -0,8\)
b) Ta có : \(\frac{2020}{2019}=1+\frac{1}{2019}\)
\(\frac{2021}{2020}=1+\frac{1}{2020}\)
Vì \(\frac{1}{2019}>\frac{1}{2020}\)nên \(\frac{2020}{2019}>\frac{2021}{2020}\)
c) \(\frac{10^{2018}+1}{10^{2019}+1}=\frac{10\left(10^{2018}+1\right)}{10^{2019}+1}=\frac{10^{2019}+10}{10^{2019}+1}=\frac{10^{2019}+1+9}{10^{2019}+1}=1+\frac{9}{10^{2019}+1}\)(1)
\(\frac{10^{2019}+1}{10^{2020}+1}=\frac{10\left(10^{2019}+1\right)}{10^{2020}+1}=\frac{10^{2020}+10}{10^{2020}+1}=\frac{10^{2020}+1+9}{10^{2020}+1}=1+\frac{9}{10^{2020}+1}\)(2)
Đến đây tự so sánh rồi nhé
